Coordinates and Geometry

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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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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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.

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.

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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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.

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 $r = \sqrt{\frac{D^2}{4} + \frac{E^2}{4} - F}$ 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 $\sqrt{\frac{D^2 + E^2}{4} - F}$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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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.

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

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

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:

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

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:

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

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 $\frac{|ax_0 + by_0 + c| \cdot |c|}{2|ab|}$. Therefore:

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

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:

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

Worked example

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

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

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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. $\blacksquare$

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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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 $F$.

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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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.

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 $\cos\theta = \frac{2}{3}$.

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

The two points are:

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

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

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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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:

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

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

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

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

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 $= \frac{-3 + 2\sqrt{2} + 3}{1 - 2} = \frac{2\sqrt{2}}{-1} = -2\sqrt{2}$.

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

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

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

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

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 $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$ 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 $AB$:

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

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

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

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

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 $3x + 6y - 8 = 0$):

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

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

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

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

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 $\theta = \frac{\pi}{3}$.

Solution

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

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

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

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 $\theta = \frac{\pi}{3}$: $x = 1 + 5\cos\frac{\pi}{3} = 1 + \frac{5}{2} = \frac{7}{2}$ $y = -2 + 5\sin\frac{\pi}{3} = -2 + \frac{5\sqrt{3}}{2}$.

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

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

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

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

Using point-slope form:

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

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

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

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

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 $\frac{|36 - 10|}{13} = \frac{26}{13} = 2 = r$.

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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