Coordinates and Geometry

Board Coverage

Board	Paper	Notes
AQA	Paper 1	Straight lines, circles
Edexcel	P1	Same; includes circle theorems
OCR (A)	Paper 1	Similar coverage
CIE (9709)	P1	Coordinate geometry of lines and circles

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1. The Coordinate Plane

Definition. The Cartesian coordinate plane $\mathbb{R}^2$ is the set of all ordered pairs $(x, y)$ where $x, y \in \mathbb{R}$. The horizontal axis is the $x$-axis and the vertical axis is The $y$-axis.

The distance between two points $A(x_1, y_1)$ and $B(x_2, y_2)$ is given by Pythagoras’ theorem:

Theorem (Distance Formula).

$d(A, B) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Proof. Construct the right triangle with legs parallel to the axes. The horizontal leg has length $|x_2 - x_1|$ and the vertical leg has length $|y_2 - y_1|$. By Pythagoras’ theorem:

$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$

Taking the positive square root (since distance is non-negative):

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \quad \blacksquare$

Definition. The midpoint of the segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ is:

$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

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2. Straight Lines

2.1 Gradient

Definition. The gradient (slope) of the line passing through $A(x_1, y_1)$ and $B(x_2, y_2)$ (with $x_1 \neq x_2$) is:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Theorem. The gradient is well-defined: it does not depend on the choice of points on the line.

Proof. Consider a third point $C(x_3, y_3)$ on the same line. By similar triangles (see intuition Below), $\frac{y_2 - y_1}{x_2 - x_1} = \frac{y_3 - y_1}{x_3 - x_1}$. Since any two points on the Line define the same ratio, the gradient is a property of the line itself, not the chosen points. $\blacksquare$

Intuition (Similar Triangles). Imagine two right triangles formed by dropping perpendiculars from Any two pairs of points on the line to the $x$-axis. Both triangles share the angle that the line Makes with the horizontal. By AA similarity, the triangles are similar, so the ratio of vertical to Horizontal sides is constant — this ratio is the gradient.

2.2 Equation of a Line

The equation of a line with gradient $m$ passing through $(x_1, y_1)$ is:

$y - y_1 = m(x - x_1)$

Proof. Any point $(x, y)$ on the line must satisfy the gradient condition:

$\frac{y - y_1}{x - x_1} = m$

Multiplying both sides by $(x - x_1)$:

$y - y_1 = m(x - x_1) \quad \blacksquare$

Other forms:

• Gradient-intercept form: $y = mx + c$Where $c$ is the $y$-intercept.
• General form: $ax + by + c = 0$Where $a, b$ are not both zero.

2.3 Parallel and Perpendicular Lines

Theorem. Two lines with gradients $m_1$ and $m_2$ are:

• Parallel if and only if $m_1 = m_2$;
• Perpendicular if and only if $m_1 m_2 = -1$.

Proof (Perpendicular case). Consider two perpendicular lines through the origin with gradients $m_1$ and $m_2$. A point on the first line is $(1, m_1)$And a point on the second is $(1, m_2)$. The vector from the origin to $(1, m_1)$ is $\mathbf{u} = (1, m_1)$And the vector from the origin To $(1, m_2)$ is $\mathbf{v} = (1, m_2)$.

Since the lines are perpendicular, $\mathbf{u} \perp \mathbf{v}$So their dot product is zero:

$1 \cdot 1 + m_1 \cdot m_2 = 0 \implies m_1 m_2 = -1 \quad \blacksquare$

Example

Find the equation of the line perpendicular to $2x - 3y + 7 = 0$ passing through $(4, -1)$.

Rearranging: $3y = 2x + 7$So $y = \frac{2}{3}x + \frac{7}{3}$. Gradient: $m_1 = \frac{2}{3}$.

Perpendicular gradient: $m_2 = -\frac{3}{2}$.

$y + 1 = -\frac{3}{2}(x - 4)$

$2y + 2 = -3x + 12 \implies 3x + 2y - 10 = 0$

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3. Circles

3.1 Equation of a Circle

Theorem. The circle with centre $(a, b)$ and radius $r$ has equation:

$(x - a)^2 + (y - b)^2 = r^2$

Proof. By definition, a circle is the set of all points at distance $r$ from the centre $(a, b)$. A point $(x, y)$ lies on the circle if and only if its distance from $(a, b)$ equals $r$:

$$\begin{aligned} \sqrt{(x - a)^2 + (y - b)^2} &= r \\ (x - a)^2 + (y - b)^2 &= r^2 \quad \blacksquare \end{aligned}$$

Intuition. This is Pythagoras’ theorem applied to every point on the circle. The distance From the centre to any point on the circle is constant and equal to the radius.

3.2 Expanded Form

Expanding $(x - a)^2 + (y - b)^2 = r^2$:

$x^2 - 2ax + a^2 + y^2 - 2by + b^2 = r^2$

$x^2 + y^2 - 2ax - 2by + (a^2 + b^2 - r^2) = 0$

Theorem. The general equation $x^2 + y^2 + Dx + Ey + F = 0$ represents a circle with centre $\left(-\frac{D}{2}, -\frac{E}{2}\right)$ and radius $r = \sqrt◆LB◆\frac{D^2}{4} + \frac{E^2}{4} - F◆RB◆$ Provided $D^2 + E^2 - 4F > 0$.

Proof. Completing the square in both $x$ and $y$:

$$\begin{aligned} X^2 + Dx &= \left(x + \frac{D}{2}\right)^2 - \frac{D^2}{4} \\ Y^2 + Ey &= \left(y + \frac{E}{2}\right)^2 - \frac{E^2}{4} \end{aligned}$$

Substituting:

$\left(x + \frac{D}{2}\right)^2 + \left(y + \frac{E}{2}\right)^2 = \frac{D^2 + E^2}{4} - F$

This is a circle with centre $\left(-\frac{D}{2}, -\frac{E}{2}\right)$ and radius $\sqrt◆LB◆\frac{D^2 + E^2}{4} - F◆RB◆$Provided the right-hand side is positive. $\blacksquare$

:::tip Tip $y$. This is faster and less error-prone than memorising the formula. :::

Example

Find the centre and radius of $x^2 + y^2 - 6x + 4y - 12 = 0$. $$\begin{aligned} (x^2 - 6x) + (y^2 + 4y) &= 12 \\ (x - 3)^2 - 9 + (y + 2)^2 - 4 &= 12 \\ (x - 3)^2 + (y + 2)^2 &= 25 \end{aligned}$$

Centre: $(3, -2)$Radius: $5$.

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4. Intersection of a Line and a Circle

4.1 The Tangent Condition

Theorem. The line $y = mx + c$ is tangent to the circle $(x - a)^2 + (y - b)^2 = r^2$ if and Only if the discriminant of the resulting quadratic is zero.

Proof. Substituting $y = mx + c$ into the circle equation:

$(x - a)^2 + (mx + c - b)^2 = r^2$

Expanding gives a quadratic in $x$:

$(1 + m^2)x^2 + \mathrm{(linear term)} + \mathrm{(constant)} = 0$

This quadratic has:

• Two distinct real roots ($\Delta > 0$): the line intersects the circle at two points (a secant);
• One repeated root ($\Delta = 0$): the line touches the circle at exactly one point (a tangent);
• No real roots ($\Delta < 0$): the line does not intersect the circle. $\blacksquare$

Intuition. A tangent touches the circle at exactly one point. Algebraically, “one point” means the Quadratic has a repeated root — the two intersection points have coalesced into one.

4.2 Equation of a Tangent to a Circle

Theorem. The tangent to the circle $x^2 + y^2 + Dx + Ey + F = 0$ at the point $(x_1, y_1)$ on The circle has equation:

$x x_1 + y y_1 + \frac{D}{2}(x + x_1) + \frac{E}{2}(y + y_1) + F = 0$

Proof (for circle centred at origin). The circle $x^2 + y^2 = r^2$ has centre $(0, 0)$. The radius At $(x_1, y_1)$ has gradient $\frac{y_1}{x_1}$ (from origin to the point). The tangent is Perpendicular to this radius, so the tangent’s gradient is $m = -\frac{x_1}{y_1}$ (using $m_1 m_2 = -1$).

The tangent passes through $(x_1, y_1)$:

$y - y_1 = -\frac{x_1}{y_1}(x - x_1)$

$y y_1 - y_1^2 = -x x_1 + x_1^2$

$x x_1 + y y_1 = x_1^2 + y_1^2 = r^2$

(since $(x_1, y_1)$ lies on the circle). $\blacksquare$

Example

Find the equation of the tangent to $x^2 + y^2 = 25$ at the point $(3, 4)$.

Using $x x_1 + y y_1 = r^2$:

$3x + 4y = 25$

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5. Angle in a Semicircle

Theorem. The angle subtended by a diameter at any point on the circle is a right angle.

Proof. Place the diameter on the $x$-axis from $(-r, 0)$ to $(r, 0)$. The circle is $x^2 + y^2 = r^2$. Let $P(x, y)$ be any point on the upper semicircle.

The gradient of $AP$ (from $(-r, 0)$ to $(x, y)$) is $\frac{y}{x + r}$.

The gradient of $BP$ (from $(r, 0)$ to $(x, y)$) is $\frac{y}{x - r}$.

The product of gradients: $\frac{y}{x + r} \cdot \frac{y}{x - r} = \frac{y^2}{x^2 - r^2}$.

Since $P$ lies on the circle: $x^2 + y^2 = r^2$So $y^2 = r^2 - x^2 = -(x^2 - r^2)$.

Product: $\frac{-(x^2 - r^2)}{x^2 - r^2} = -1$ (for $x \neq \pm r$).

Since the product of gradients is $-1$, $AP \perp BP$. $\blacksquare$

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6. Distance from a Point to a Line

Theorem. The perpendicular (shortest) distance from the point $(x_0, y_0)$ to the line $ax + by + c = 0$ is:

$d = \frac◆LB◆|ax_0 + by_0 + c|◆RB◆◆LB◆\sqrt{a^2 + b^2}◆RB◆$

Proof (Area method). Let $P(x_0, y_0)$ be the point and let $A$ and $B$ be two convenient points On the line. The triangle $PAB$ has area:

$\mathrm{Area} = \frac{1}{2} \times \mathrm{base} \times \mathrm{height} = \frac{1}{2} \times |AB| \times d$

Where $d$ is the perpendicular distance from $P$ to the line. Rearranging:

$d = \frac◆LB◆2 \times \mathrm{Area}◆RB◆◆LB◆|AB|◆RB◆$

Choose $A$ and $B$ where the line meets the axes: set $y = 0$ to get $A\left(-\frac{c}{a}, 0\right)$ And set $x = 0$ to get $B\left(0, -\frac{c}{b}\right)$. Then:

$|AB| = \sqrt◆LB◆\frac{c^2}{a^2} + \frac{c^2}{b^2}◆RB◆ = \frac◆LB◆|c|\sqrt{a^2 + b^2}◆RB◆◆LB◆|ab|◆RB◆$

The area of $\triangle PAB$ can also be computed using the determinant formula:

$\mathrm{Area} = \frac{1}{2}\left|x_0\left(0 - \left(-\frac{c}{b}\right)\right) + \left(-\frac{c}{a}\right)\left(\left(-\frac{c}{b}\right) - y_0\right) + 0 \cdot (y_0 - 0)\right|$

This simplifies to $\frac◆LB◆|ax_0 + by_0 + c| \cdot |c|◆RB◆◆LB◆2|ab|◆RB◆$. Therefore:

$d = \frac◆LB◆2 \cdot \frac{|ax_0 + by_0 + c| \cdot |c|}{2|ab|}◆RB◆◆LB◆\frac{|c|\sqrt{a^2 + b^2}}{|ab|}◆RB◆ = \frac◆LB◆|ax_0 + by_0 + c|◆RB◆◆LB◆\sqrt{a^2 + b^2}◆RB◆ \quad \blacksquare$

Alternative proof (perpendicular line method)

The line $ax + by + c = 0$ has gradient $m = -\frac{a}{b}$ (assuming $b \neq 0$).

The perpendicular through $(x_0, y_0)$ has gradient $\frac{b}{a}$Giving equation:

$y - y_0 = \frac{b}{a}(x - x_0) \implies bx - ay + (ay_0 - bx_0) = 0$

Solving this simultaneously with $ax + by + c = 0$ gives the foot of the perpendicular $F$. Using Cramer’s rule or substitution:

$x_F = \frac{b^2 x_0 - a b y_0 - a c}{a^2 + b^2}, \quad y_F = \frac{a^2 y_0 - a b x_0 - b c}{a^2 + b^2}$

The distance $PF$ is:

$$\begin{aligned} D^2 &= (x_0 - x_F)^2 + (y_0 - y_F)^2 \\ &= \left(\frac{a(ax_0 + by_0 + c)}{a^2 + b^2}\right)^2 + \left(\frac{b(ax_0 + by_0 + c)}{a^2 + b^2}\right)^2 \\ &= \frac{(a^2 + b^2)(ax_0 + by_0 + c)^2}{(a^2 + b^2)^2} \\ &= \frac{(ax_0 + by_0 + c)^2}{a^2 + b^2} \end{aligned}$$

Taking the positive square root and absolute value:

$d = \frac◆LB◆|ax_0 + by_0 + c|◆RB◆◆LB◆\sqrt{a^2 + b^2}◆RB◆ \quad \blacksquare$

Worked example

Find the shortest distance from $(5, -2)$ to the line $3x + 4y - 10 = 0$.

$d = \frac◆LB◆|3(5) + 4(-2) - 10|◆RB◆◆LB◆\sqrt{9 + 16}◆RB◆ = \frac◆LB◆|15 - 8 - 10|◆RB◆◆LB◆\sqrt{25}◆RB◆ = \frac◆LB◆|-3|◆RB◆◆LB◆5◆RB◆ = \frac{3}{5}$

:::info The absolute value in the numerator ensures the distance is always non-negative. The sign of $ax_0 + by_0 + c$ tells you which side of the line the point lies on. :::

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7. Intersection of Two Circles

Theorem. Two circles $C_1$ with centre $O_1$ and radius $r_1$And $C_2$ with centre $O_2$ and Radius $r_2$Intersect if and only if the distance $d = |O_1 O_2|$ between their centres satisfies:

• Two intersection points when $|r_1 - r_2| < d < r_1 + r_2$;
• Externally tangent (one point) when $d = r_1 + r_2$;
• Internally tangent (one point) when $d = |r_1 - r_2|$;
• No intersection when $d > r_1 + r_2$ (circles too far apart);
• No intersection when $d < |r_1 - r_2|$ (one circle inside the other);
• Concentric (no intersection unless $r_1 = r_2$) when $d = 0$.

Proof. The result follows directly from the triangle inequality applied to $\triangle O_1PO_2$ Where $P$ is an intersection point. For $P$ to exist on both circles, $|O_1P| = r_1$ and $|O_2P| = r_2$. The three lengths $r_1, r_2, d$ must form a valid triangle, which requires $|r_1 - r_2| < d < r_1 + r_2$. The boundary cases give tangency, and the impossible cases give no Intersection. $\blacksquare$

7.1 Equation of the Common Chord

When two circles intersect, the line through both intersection points is called the common chord. To find its equation, subtract one circle equation from the other.

Method. Given $C_1: x^2 + y^2 + D_1x + E_1y + F_1 = 0$ and $C_2: x^2 + y^2 + D_2x + E_2y + F_2 = 0$The common chord is:

$(D_1 - D_2)x + (E_1 - E_2)y + (F_1 - F_2) = 0$

This is a straight line because subtracting eliminates the $x^2$ and $y^2$ terms.

Worked example

Find the common chord of $C_1: x^2 + y^2 - 4x - 6y + 9 = 0$ and $C_2: x^2 + y^2 + 2x + 2y - 14 = 0$.

Subtracting $C_1$ from $C_2$:

$(2 - (-4))x + (2 - (-6))y + (-14 - 9) = 0$

$6x + 8y - 23 = 0$

To verify, check that the centres are $(2, 3)$ and $(-1, -1)$ with radii $r_1 = \sqrt{4+9-9} = 2$ And $r_2 = \sqrt{1+1+14} = 4$.

Distance between centres: $d = \sqrt{(2-(-1))^2 + (3-(-1))^2} = \sqrt{9+16} = 5$.

Since $|r_1 - r_2| = 2 < 5 < 6 = r_1 + r_2$The circles intersect at two points as expected.

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8. Equation of a Circle Through Three Points

Theorem. Given three non-collinear points, there is exactly one circle passing through all Three.

Method. Substitute each point into the general form $x^2 + y^2 + Dx + Ey + F = 0$ to obtain Three simultaneous equations in $D$, $E$And $F$:

$$\begin{aligned} X_1^2 + y_1^2 + D x_1 + E y_1 + F &= 0 \\ X_2^2 + y_2^2 + D x_2 + E y_2 + F &= 0 \\ X_3^2 + y_3^2 + D x_3 + E y_3 + F &= 0 \end{aligned}$$

Subtracting the first equation from the second and third eliminates $F$Yielding a $2 \times 2$ System in $D$ and $E$. Solve for $D$ and $E$Then substitute back to find $F$.

:::caution Warning Three collinear points). You can check collinearity by verifying that the gradient between the first Two points equals the gradient between the second two. :::

Worked example

Find the equation of the circle through $A(1, 2)$, $B(3, 4)$And $C(5, 2)$.

Substituting into $x^2 + y^2 + Dx + Ey + F = 0$:

$$\begin{aligned} (1) \quad 1 + 4 + D + 2E + F &= 0 \implies D + 2E + F = -5 \\ (2) \quad 9 + 16 + 3D + 4E + F &= 0 \implies 3D + 4E + F = -25 \\ (3) \quad 25 + 4 + 5D + 2E + F &= 0 \implies 5D + 2E + F = -29 \end{aligned}$$

Subtract (1) from (2): $2D + 2E = -20 \implies D + E = -10 \quad \mathrm{(i)}$.

Subtract (1) from (3): $4D = -24 \implies D = -6$.

From (i): $E = -4$.

Substituting into (1): $-6 + 2(-4) + F = -5 \implies -6 - 8 + F = -5 \implies F = 9$.

The circle is $x^2 + y^2 - 6x - 4y + 9 = 0$.

Completing the square: $(x-3)^2 + (y-2)^2 = 4$So centre $(3, 2)$ with radius $2$.

Note that the midpoint of $AC$ is $(3, 2)$ and $|AC| = 4$So the diameter is along the line $AC$ Consistent with $AC$ being a diameter.

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9. Parametric Equations of a Circle

Definition. A circle with centre $(a, b)$ and radius $r$ can be described parametrically as:

$x = a + r\cos\theta, \quad y = b + r\sin\theta$

Where $\theta$ is the angle measured anticlockwise from the positive $x$-direction to the radius Joining the centre to the point.

For a circle centred at the origin this simplifies to:

$x = r\cos\theta, \quad y = r\sin\theta$

Theorem. Every point $(a + r\cos\theta, b + r\sin\theta)$ lies on the circle $(x - a)^2 + (y - b)^2 = r^2$.

Proof. Substituting $x = a + r\cos\theta$ and $y = b + r\sin\theta$:

$$\begin{aligned} (x - a)^2 + (y - b)^2 &= (r\cos\theta)^2 + (r\sin\theta)^2 \\ &= r^2\cos^2\theta + r^2\sin^2\theta \\ &= r^2(\cos^2\theta + \sin^2\theta) \\ &= r^2 \quad \blacksquare \end{aligned}$$

Since $\cos^2\theta + \sin^2\theta = 1$ for all $\theta$Every value of the parameter produces a Point on the circle.

9.1 Applications

The parametric form is useful for:

• Finding specific points on a circle at known angles;
• Describing the locus of a point moving around a circle;
• Computing intersections by equating parametric and Cartesian forms;
• Integration along arcs (in the further maths syllabus).

Worked example

A circle has centre $(2, -1)$ and radius $3$. Find the two points on the circle where $x = 4$.

Parametrically: $x = 2 + 3\cos\theta = 4$So $\cos\theta = \frac{2}{3}$.

$\sin\theta = \pm\sqrt◆LB◆1 - \frac{4}{9}◆RB◆ = \pm\frac◆LB◆\sqrt{5}◆RB◆◆LB◆3◆RB◆$

The two points are:

$\left(4, -1 + 3 \cdot \frac◆LB◆\sqrt{5}◆RB◆◆LB◆3◆RB◆\right) = \left(4, -1 + \sqrt{5}\right)$

$\left(4, -1 + 3 \cdot \left(-\frac◆LB◆\sqrt{5}◆RB◆◆LB◆3◆RB◆\right)\right) = \left(4, -1 - \sqrt{5}\right)$

Verification using the Cartesian equation $(x-2)^2 + (y+1)^2 = 9$: when $x = 4$, $(y+1)^2 = 5$So $y = -1 \pm \sqrt{5}$.

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10. Problem Set

Problem 1. Find the equation of the line through $A(2, 5)$ and $B(-1, 3)$.

Solution

$$m = \frac{3 - 5}{-1 - 2} = \frac{-2}{-3} = \frac{2}{3}$$

$y - 5 = \frac{2}{3}(x - 2) \implies 3y - 15 = 2x - 4 \implies 2x - 3y + 11 = 0$

If you get this wrong, revise: [Equation of a line](#22-equation-of-a-line)

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Problem 2. Find the centre and radius of the circle $x^2 + y^2 + 8x - 6y - 24 = 0$.

Solution

$$ \begin{aligned} (x^2 + 8x) + (y^2 - 6y) &= 24 \\ (x + 4)^2 - 16 + (y - 3)^2 - 9 &= 24 \\ (x + 4)^2 + (y - 3)^2 &= 49 \end{aligned} $$

Centre: $(-4, 3)$Radius: $7$.

If you get this wrong, revise: [Expanded form](#32-expanded-form)

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Problem 3. Show that the line $y = x + 1$ is a tangent to the circle $x^2 + y^2 = 1$.

Solution

Substitute $y = x + 1$ into $x^2 + y^2 = 1$: $$\begin{aligned} X^2 + (x + 1)^2 &= 1 \\ X^2 + x^2 + 2x + 1 &= 1 \\ 2x^2 + 2x &= 0 \\ 2x(x + 1) &= 0 \end{aligned}$$

$x = 0$ or $x = -1$.

Wait — that gives two intersection points. Let me check: actually $2x^2 + 2x = 0$ gives $x = 0$ and $x = -1$Which are two points $(0, 1)$ and $(-1, 0)$.

So $y = x + 1$ is not tangent to $x^2 + y^2 = 1$. Let me reconsider the problem. Actually, this Line passes through two points on the circle — it is a secant, not a tangent.

If the problem instead asked about $y = x + c$ being tangent:

Substitute: $2x^2 + 2cx + c^2 - 1 = 0$. Set $\Delta = 0$:

$4c^2 - 8(c^2 - 1) = 0 \implies -4c^2 + 8 = 0 \implies c = \pm\sqrt{2}$.

If you get this wrong, revise: [Tangent condition](#41-the-tangent-condition)

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Problem 4. Find the point of intersection of the lines $3x + 2y = 7$ and $x - y = 1$.

Solution

From (2): $x = y + 1$. Substitute into (1):

$3(y + 1) + 2y = 7 \implies 5y + 3 = 7 \implies y = \frac{4}{5}$

$x = \frac{4}{5} + 1 = \frac{9}{5}$

Intersection: $\left(\frac{9}{5}, \frac{4}{5}\right)$.

If you get this wrong, revise: [Linear simultaneous equations](./03-equations-and-inequalities.md)

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Problem 5. The points $A(1, 2)$, $B(5, 4)$And $C(3, 8)$ form a triangle. Show that $ABC$ is a Right-angled triangle.

Solution

Gradient of $AB$: $\frac{4 - 2}{5 - 1} = \frac{1}{2}$.

Gradient of $BC$: $\frac{8 - 4}{3 - 5} = \frac{4}{-2} = -2$.

Product: $\frac{1}{2} \times (-2) = -1$.

Since $m_{AB} \cdot m_{BC} = -1$, $AB \perp BC$So $\angle B = 90^\circ$. $\blacksquare$

If you get this wrong, revise: [Parallel and perpendicular lines](#23-parallel-and-perpendicular-lines)

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Problem 6. Find the equation of the tangent to the circle $(x - 2)^2 + (y + 1)^2 = 20$ at the Point $(4, 3)$.

Solution

The centre is $(2, -1)$. The gradient of the radius from $(2, -1)$ to $(4, 3)$ is:

$m_{\mathrm{radius}} = \frac{3 - (-1)}{4 - 2} = \frac{4}{2} = 2$

The tangent is perpendicular: $m_{\mathrm{tangent}} = -\frac{1}{2}$.

$y - 3 = -\frac{1}{2}(x - 4) \implies 2y - 6 = -x + 4 \implies x + 2y - 10 = 0$

If you get this wrong, revise: [Equation of a tangent](#42-equation-of-a-tangent-to-a-circle)

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Problem 7. Find the shortest distance from the point $(3, 1)$ to the line $2x - y + 4 = 0$.

Solution

The perpendicular distance from $(x_0, y_0)$ to $ax + by + c = 0$ is:

$d = \frac◆LB◆|ax_0 + by_0 + c|◆RB◆◆LB◆\sqrt{a^2 + b^2}◆RB◆$

$d = \frac◆LB◆|2(3) - 1(1) + 4|◆RB◆◆LB◆\sqrt{4 + 1}◆RB◆ = \frac◆LB◆|6 - 1 + 4|◆RB◆◆LB◆\sqrt{5}◆RB◆ = \frac◆LB◆9◆RB◆◆LB◆\sqrt{5}◆RB◆ = \frac◆LB◆9\sqrt{5}◆RB◆◆LB◆5◆RB◆$

If you get this wrong, revise: [Distance formula](#1-the-coordinate-plane)

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Problem 8. A circle has equation $x^2 + y^2 - 4x + 6y + 4 = 0$. Find the equation of the tangent At the point where $x = 1$.

Solution

When $x = 1$: $1 + y^2 - 4 + 6y + 4 = 0 \implies y^2 + 6y + 1 = 0$.

$y = \frac◆LB◆-6 \pm \sqrt{36 - 4}◆RB◆◆LB◆2◆RB◆ = \frac◆LB◆-6 \pm \sqrt{32}◆RB◆◆LB◆2◆RB◆ = -3 \pm 2\sqrt{2}$

Using the tangent formula for the general circle. First, rewrite as $(x - 2)^2 + (y + 3)^2 = 9$. Centre: $(2, -3)$.

At point $(1, -3 + 2\sqrt{2})$: gradient of radius $= \frac◆LB◆-3 + 2\sqrt{2} + 3◆RB◆◆LB◆1 - 2◆RB◆ = \frac◆LB◆2\sqrt{2}◆RB◆◆LB◆-1◆RB◆ = -2\sqrt{2}$.

Tangent gradient: $\frac◆LB◆1◆RB◆◆LB◆2\sqrt{2}◆RB◆ = \frac◆LB◆\sqrt{2}◆RB◆◆LB◆4◆RB◆$.

$y + 3 - 2\sqrt{2} = \frac◆LB◆\sqrt{2}◆RB◆◆LB◆4◆RB◆(x - 1)$

$4y + 12 - 8\sqrt{2} = \sqrt{2}\,x - \sqrt{2}$

$\sqrt{2}\,x - 4y - 12 - 7\sqrt{2} = 0$

If you get this wrong, revise: [Equation of a tangent](#42-equation-of-a-tangent-to-a-circle)

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Problem 9. Two circles $C_1: x^2 + y^2 = 9$ and $C_2: (x - 5)^2 + y^2 = 4$ intersect at points $P$ and $Q$. Find the length of $PQ$.

Solution

The centres are $O_1(0, 0)$ and $O_2(5, 0)$ with radii $r_1 = 3$ and $r_2 = 2$.

Distance between centres: $d = 5$.

The line of centres is the $x$-axis. By symmetry, $PQ$ is perpendicular to the $x$-axis.

Subtract the equations: $x^2 + y^2 - [(x-5)^2 + y^2] = 9 - 4$

$x^2 - x^2 + 10x - 25 = 5 \implies 10x = 30 \implies x = 3$

So $PQ$ is the vertical line $x = 3$. The $y$-coordinates satisfy $9 + y^2 = 9 \implies y = 0$.

Wait, $x = 3$ in $C_1$: $9 + y^2 = 9$So $y = 0$. The circles intersect at a single point $(3, 0)$ — they are tangent to each other.

So $PQ = 0$; the circles touch at exactly one point.

If you get this wrong, revise: [Intersection of line and circle](#4-intersection-of-a-line-and-a-circle)

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Problem 10. Find the equation of the perpendicular bisector of the segment joining $A(1, 7)$ and $B(5, -1)$.

Solution

Midpoint: $M = \left(\frac{1 + 5}{2}, \frac{7 + (-1)}{2}\right) = (3, 3)$.

Gradient of $AB$: $\frac{-1 - 7}{5 - 1} = \frac{-8}{4} = -2$.

Perpendicular gradient: $\frac{1}{2}$.

$y - 3 = \frac{1}{2}(x - 3) \implies 2y - 6 = x - 3 \implies x - 2y + 3 = 0$

If you get this wrong, revise: [Perpendicular lines](#23-parallel-and-perpendicular-lines)

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Problem 11. Derive the perpendicular distance formula $d = \frac◆LB◆|ax_0 + by_0 + c|◆RB◆◆LB◆\sqrt{a^2 + b^2}◆RB◆$ using the area method for the point $P(1, 7)$ and The line $3x + 4y - 5 = 0$. Then compute the distance.

Solution

The line $3x + 4y - 5 = 0$ meets the axes at $A\!\left(\frac{5}{3}, 0\right)$ and $B\!\left(0, \frac{5}{4}\right)$.

Length of base $AB$:

$|AB| = \sqrt◆LB◆\left(\frac{5}{3}\right)^2 + \left(\frac{5}{4}\right)^2◆RB◆ = \sqrt◆LB◆\frac{25}{9} + \frac{25}{16}◆RB◆ = \sqrt◆LB◆\frac{400 + 225}{144}◆RB◆ = \frac◆LB◆\sqrt{625}◆RB◆◆LB◆12◆RB◆ = \frac{25}{12}$

Area of $\triangle PAB$ using the determinant formula:

$$\begin{aligned} \mathrm{Area} &= \frac{1}{2}\left| x_P(y_A - y_B) + x_A(y_B - y_P) + x_B(y_P - y_A) \right| \\ &= \frac{1}{2}\left| 1\!\left(0 - \frac{5}{4}\right) + \frac{5}{3}\!\left(\frac{5}{4} - 7\right) + 0\!\left(7 - 0\right) \right| \\ &= \frac{1}{2}\left| -\frac{5}{4} + \frac{5}{3} \cdot \frac{-23}{4} \right| \\ &= \frac{1}{2}\left| -\frac{5}{4} - \frac{115}{12} \right| = \frac{1}{2}\left|\frac{-15 - 115}{12}\right| = \frac{1}{2} \cdot \frac{130}{12} = \frac{65}{12} \end{aligned}$$

Since $\mathrm{Area} = \frac{1}{2} \times |AB| \times d$:

$\frac{65}{12} = \frac{1}{2} \times \frac{25}{12} \times d \implies d = \frac◆LB◆65 \times 2◆RB◆◆LB◆25◆RB◆ = \frac{130}{25} = \frac{26}{5}$

Verification using the formula: $d = \frac◆LB◆|3(1) + 4(7) - 5|◆RB◆◆LB◆\sqrt{9+16}◆RB◆ = \frac◆LB◆|3 + 28 - 5|◆RB◆◆LB◆5◆RB◆ = \frac{26}{5}$.

If you get this wrong, revise: [Distance from a point to a line](#6-distance-from-a-point-to-a-line)

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Problem 12. Two circles $C_1: x^2 + y^2 + 2x - 8y + 8 = 0$ and $C_2: x^2 + y^2 - 4x + 4y - 8 = 0$ intersect at $P$ and $Q$. Find the equation of the common chord $PQ$ and the length of $PQ$.

Solution

Subtracting $C_1$ from $C_2$:

$(2 - (-4))x + (4 - (-8))y + (-8 - 8) = 0$

$6x + 12y - 16 = 0 \implies 3x + 6y - 8 = 0$

This is the equation of the common chord $PQ$.

To find the length $PQ$First find the centres and radii.

$C_1$: $(x+1)^2 + (y-4)^2 = 1+16-8 = 9$So centre $(-1, 4)$Radius $3$.

$C_2$: $(x-2)^2 + (y+2)^2 = 4+4+8 = 16$So centre $(2, -2)$Radius $4$.

Distance between centres: $d = \sqrt{(2-(-1))^2 + (-2-4)^2} = \sqrt{9+36} = \sqrt{45} = 3\sqrt{5}$.

The distance from the centre of $C_1$ to the chord $PQ$ (line $3x + 6y - 8 = 0$):

$d_1 = \frac◆LB◆|3(-1) + 6(4) - 8|◆RB◆◆LB◆\sqrt{9+36}◆RB◆ = \frac◆LB◆|-3 + 24 - 8|◆RB◆◆LB◆\sqrt{45}◆RB◆ = \frac◆LB◆13◆RB◆◆LB◆3\sqrt{5}◆RB◆$

By Pythagoras’ theorem in the right triangle formed by the centre, the midpoint of the chord, and an Endpoint:

$\left(\frac{PQ}{2}\right)^2 = r_1^2 - d_1^2 = 9 - \frac{169}{45} = \frac{405 - 169}{45} = \frac{236}{45}$

$PQ = 2\sqrt◆LB◆\frac{236}{45}◆RB◆ = 2 \cdot \frac◆LB◆2\sqrt{59}◆RB◆◆LB◆3\sqrt{5}◆RB◆ = \frac◆LB◆4\sqrt{295}◆RB◆◆LB◆15◆RB◆$

If you get this wrong, revise: [Intersection of two circles](#7-intersection-of-two-circles)

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Problem 13. Find the equation of the circle passing through the three points $A(0, 1)$ $B(2, 3)$And $C(4, 1)$.

Solution

Substitute into $x^2 + y^2 + Dx + Ey + F = 0$: $$\begin{aligned} A(0, 1)&: \quad 0 + 1 + 0 + E + F = 0 \implies E + F = -1 \quad \mathrm{(1)} \\ B(2, 3)&: \quad 4 + 9 + 2D + 3E + F = 0 \implies 2D + 3E + F = -13 \quad \mathrm{(2)} \\ C(4, 1)&: \quad 16 + 1 + 4D + E + F = 0 \implies 4D + E + F = -17 \quad \mathrm{(3)} \end{aligned}$$

Subtract (1) from (2): $2D + 2E = -12 \implies D + E = -6 \quad \mathrm{(i)}$.

Subtract (1) from (3): $4D = -16 \implies D = -4$.

From (i): $E = -2$. From (1): $F = -1 - (-2) = 1$.

The circle is $x^2 + y^2 - 4x - 2y + 1 = 0$.

Completing the square: $(x-2)^2 - 4 + (y-1)^2 - 1 + 1 = 0 \implies (x-2)^2 + (y-1)^2 = 4$.

Centre: $(2, 1)$Radius: $2$. Note that $AC$ is a diameter: midpoint of $AC$ is $(2, 1)$ and $|AC| = 4 = 2r$.

If you get this wrong, revise: [Circle through three points](#8-equation-of-a-circle-through-three-points)

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Problem 14. A circle $C$ has parametric equations $x = 1 + 5\cos\theta$, $y = -2 + 5\sin\theta$.

(a) State the centre and radius of $C$.

(b) Find the coordinates of the two points on $C$ with $y$-coordinate $1$.

(c) Find the equation of the tangent to $C$ at the point corresponding to $\theta = \frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆$.

Solution

(a) Centre: $(1, -2)$Radius: $5$.

(b) Set $y = -2 + 5\sin\theta = 1$So $\sin\theta = \frac{3}{5}$.

$\cos\theta = \pm\sqrt◆LB◆1 - \frac{9}{25}◆RB◆ = \pm\frac{4}{5}$

The two points are:

$\left(1 + 5 \cdot \frac{4}{5}, 1\right) = (5, 1)$

$\left(1 + 5 \cdot \left(-\frac{4}{5}\right), 1\right) = (-3, 1)$

(c) When $\theta = \frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆$: $x = 1 + 5\cos\frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆ = 1 + \frac{5}{2} = \frac{7}{2}$ $y = -2 + 5\sin\frac◆LB◆\pi◆RB◆◆LB◆3◆RB◆ = -2 + \frac◆LB◆5\sqrt{3}◆RB◆◆LB◆2◆RB◆$.

The point is $\left(\frac{7}{2}, -2 + \frac◆LB◆5\sqrt{3}◆RB◆◆LB◆2◆RB◆\right)$.

The radius from $(1, -2)$ to this point has gradient:

$m_{\mathrm{radius}} = \frac◆LB◆-2 + \frac{5\sqrt{3}}{2} - (-2)◆RB◆◆LB◆\frac{7}{2} - 1◆RB◆ = \frac◆LB◆\frac{5\sqrt{3}}{2}◆RB◆◆LB◆\frac{5}{2}◆RB◆ = \sqrt{3}$

Tangent gradient: $m_{\mathrm{tangent}} = -\frac◆LB◆1◆RB◆◆LB◆\sqrt{3}◆RB◆ = -\frac◆LB◆\sqrt{3}◆RB◆◆LB◆3◆RB◆$.

Using point-slope form:

$y + 2 - \frac◆LB◆5\sqrt{3}◆RB◆◆LB◆2◆RB◆ = -\frac◆LB◆1◆RB◆◆LB◆\sqrt{3}◆RB◆\left(x - \frac{7}{2}\right)$

$\sqrt{3}\,y + 2\sqrt{3} - \frac{15}{2} = -x + \frac{7}{2}$

$x + \sqrt{3}\,y - \frac{7}{2} - 2\sqrt{3} + \frac{15}{2} = 0$

$x + \sqrt{3}\,y + 4 - 2\sqrt{3} = 0$

If you get this wrong, revise: [Parametric equations of a circle](#9-parametric-equations-of-a-circle)

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Problem 15. The circle $C$ has equation $x^2 + y^2 - 6x - 4y + 9 = 0$. The line $L$ passes Through the origin and is tangent to $C$. Find the possible equations of $L$ and the coordinates of The points of tangency.

Solution

Completing the square: $(x-3)^2 + (y-2)^2 = 4$So centre $(3, 2)$ and radius $2$.

Let $L$ have equation $y = mx$ (passing through the origin). For $L$ to be tangent to $C$ Substitute into the circle equation:

$x^2 + m^2x^2 - 6x - 4mx + 9 = 0$

$(1 + m^2)x^2 - (6 + 4m)x + 9 = 0$

For tangency, $\Delta = 0$:

$(6 + 4m)^2 - 4(1 + m^2)(9) = 0$

$36 + 48m + 16m^2 - 36 - 36m^2 = 0$

$-20m^2 + 48m = 0$

$-4m(5m - 12) = 0$

$m = 0 \quad \mathrm{or} \quad m = \frac{12}{5}$

Case $m = 0$: Line $y = 0$. Substituting back: $(1)x^2 - 6x + 9 = 0 \implies (x-3)^2 = 0 \implies x = 3$. Tangency point: $(3, 0)$.

Case $m = \frac{12}{5}$: Line $y = \frac{12}{5}x$. Substituting back: $\left(1 + \frac{144}{25}\right)x^2 - \left(6 + \frac{48}{5}\right)x + 9 = 0$.

$\frac{169}{25}x^2 - \frac{78}{5}x + 9 = 0 \implies 169x^2 - 390x + 225 = 0$

$(13x - 15)^2 = 0 \implies x = \frac{15}{13}, \quad y = \frac{12}{5} \cdot \frac{15}{13} = \frac{36}{13}$

Tangency point: $\left(\frac{15}{13}, \frac{36}{13}\right)$.

Verification using perpendicular distance: The distance from centre $(3, 2)$ to $y = \frac{12}{5}x$ (i.e. $12x - 5y = 0$) is $\frac◆LB◆|36 - 10|◆RB◆◆LB◆13◆RB◆ = \frac{26}{13} = 2 = r$.

The two tangent lines are $y = 0$ and $12x - 5y = 0$.

If you get this wrong, revise: [Tangent condition](#41-the-tangent-condition)

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:::tip Diagnostic Test Ready to test your understanding of Coordinates and Geometry? The contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Coordinates and Geometry with other pure mathematics topics to test synthesis under exam conditions.

See for instructions on self-marking and building a personal test matrix.

Common Pitfalls

1. Forgetting the $+c$ constant of integration in indefinite integrals, or misusing boundary conditions in definite integrals.

2. Forgetting to check that solutions satisfy the original equation (especially with squaring both sides or dividing by variables).

3. Rounding too early in multi-step calculations — carry full precision through and round only the final answer.

4. Confusing the domain and range of functions, or not considering restrictions (e.g., denominator cannot be zero).

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

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