Coordinate Geometry Overview – A Level Maths Year 1

Coordinate geometry turns pictures into equations. Once a line or a circle has an equation, geometric questions — Do these lines cross? Is this angle a right angle? Does this line touch the circle? — become algebra questions you can answer with certainty.

Every problem in this topic rests on four facts: the gradient between two points, the distance between two points, the midpoint of two points, and the rule $m_1 m_2 = -1$ for perpendicular lines. Lines and circles are both built from these.

What is new compared with GCSE

$y - y_1 = m(x - x_1)$ and $ax + by + c = 0$ · lines in context
Circles with any centre: $(x-a)^2 + (y-b)^2 = r^2$
Completing the square to find a centre · the discriminant for tangency

Always sketch. A rough diagram with the points roughly in the right place costs ten seconds and shows you immediately whether an answer like "centre $(-3, 2)$" is plausible. This is DfE content area C · Coordinate geometry in the $(x, y)$ plane.

Word	Meaning	Example
Gradient	Change in $y$ divided by change in $x$; steepness.	$y = 3x - 1$ has gradient $3$
Intercept	Where a graph crosses an axis.	$y = 3x - 1$ crosses the $y$-axis at $(0, -1)$
Parallel	Same direction; equal gradients.	$y = 2x$ and $y = 2x + 5$
Perpendicular	At right angles; gradients multiply to $-1$.	$y = 2x$ and $y = -\tfrac12 x$
Collinear	Lying on one straight line.	$(0, 0)$, $(1, 2)$, $(3, 6)$
Perpendicular bisector	The line at right angles to $AB$ through its midpoint.	All points equidistant from $A$ and $B$
Chord	A line segment joining two points of a circle.	A diameter is the longest chord
Tangent	A line touching a circle at exactly one point.	Perpendicular to the radius there
Discriminant	$b^2 - 4ac$ for a quadratic $ax^2 + bx + c$.	$= 0$ means a repeated root (tangency)

"Exact" means no decimals. A length like $\sqrt{45}$ should be left as $3\sqrt5$, not $6.708\ldots$. Simplifying surds from Indices and Surds is used constantly here.

1 — Gradient, midpoint and distance

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$ · $\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$ · $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

2 — Equation of a line

$y - y_1 = m(x - x_1)$, then tidy into $y = mx + c$ or $ax + by + c = 0$

3 — Parallel and perpendicular

Parallel: $m_1 = m_2$ · Perpendicular: $m_1 m_2 = -1$

4 — Equation of a circle

$(x - a)^2 + (y - b)^2 = r^2$: centre $(a, b)$, radius $r$

5 — Completing the square

$x^2 + y^2 + px + qy + c = 0 \;\to\; \left(x + \tfrac{p}{2}\right)^2 + \left(y + \tfrac{q}{2}\right)^2 = \tfrac{p^2}{4} + \tfrac{q^2}{4} - c$

6 — Line meets circle

Substitute, form a quadratic, use the discriminant: $>0$ two points, $=0$ tangent, $<0$ none

Worked Example — Several techniques in one question

The circle $x^2 + y^2 - 4x + 2y - 20 = 0$ has centre $C$. The point $P(5, 3)$ lies on the circle. Find the equation of the tangent at $P$, and where it crosses the $y$-axis.

①Complete the square: $(x - 2)^2 - 4 + (y + 1)^2 - 1 - 20 = 0$, so $(x-2)^2 + (y+1)^2 = 25$. Centre $C(2, -1)$, radius $5$.

②Check $P$: $3^2 + 4^2 = 25$ ✓

③Gradient of $CP$: $\dfrac{3 - (-1)}{5 - 2} = \dfrac43$, so the tangent gradient is $-\tfrac34$.

④$y - 3 = -\tfrac34(x - 5) \;\Rightarrow\; 4y - 12 = -3x + 15 \;\Rightarrow\; 3x + 4y - 27 = 0$

⑤On the $y$-axis $x = 0$: $4y = 27$, so the tangent crosses at $\left(0, \tfrac{27}{4}\right)$.

This uses completing the square, the perpendicular gradient rule, the point–gradient form and an intercept — four techniques for one question, which is typical.

The content of DfE area C, and the page that covers it.

CStraight Lines$y - y_1 = m(x - x_1)$ and $ax + by + c = 0$; parallel and perpendicular lines; midpoints, distances and intersections; straight-line models.
CCircles$(x-a)^2 + (y-b)^2 = r^2$; completing the square; angle in a semicircle, chords and tangents; the discriminant for line–circle intersections.

Year 2 extends this topic to curves given by parametric equations.

Reading the gradient straight off $ax + by + c = 0$. The gradient of $2x + 3y = 6$ is not $2$. Rearrange: $y = -\tfrac23 x + 2$, so $m = -\tfrac23$.

Half the perpendicular rule. The perpendicular to gradient $\tfrac25$ is $-\tfrac52$: flip and change sign. $-\tfrac25$ and $\tfrac52$ are both wrong.

Wrong signs for the centre. $(x + 3)^2 + (y - 2)^2 = 16$ has centre $(-3, 2)$, not $(3, -2)$.

Giving $r^2$ as the radius. In $(x-1)^2 + y^2 = 20$ the radius is $\sqrt{20} = 2\sqrt5$, not $20$.

Subtracting the wrong way in a gradient. Keep the same order on top and bottom: $\dfrac{y_2 - y_1}{x_2 - x_1}$, never $\dfrac{y_2 - y_1}{x_1 - x_2}$.

Gradient

$\dfrac{y_2 - y_1}{x_2 - x_1}$

Line

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

Perpendicular

$m_1 m_2 = -1$

Midpoint

Average the coordinates.

Distance

Pythagoras on the changes in $x$ and $y$.

Circle

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

Tangent

Perpendicular to the radius.

Chord

Perpendicular bisector goes through the centre.

Semicircle

Angle subtended by a diameter is $90°$.

Discriminant

$b^2 - 4ac = 0$ means tangent.

Ten questions drawing on the whole Coordinate Geometry topic. Work each one through before opening the solution.

Question 1

Find the gradient of the line through $(-1, 3)$ and $(3, 11)$.

▶ Show solution

$m = \dfrac{11 - 3}{3 - (-1)} = \dfrac{8}{4} = 2$

Question 2

Find the equation of the line through $(2, 1)$ perpendicular to $y = \tfrac12 x + 3$.

▶ Show solution

Perpendicular gradient: $-2$.

$y - 1 = -2(x - 2)$, so $y = -2x + 5$.

Question 3

Find the distance between $(1, -2)$ and $(7, 6)$.

▶ Show solution

$\sqrt{6^2 + 8^2} = \sqrt{100} = 10$

Question 4

Write $4x - 6y + 3 = 0$ in the form $y = mx + c$.

▶ Show solution

$6y = 4x + 3$, so $y = \tfrac23 x + \tfrac12$.

Question 5

State the centre and the exact radius of $(x - 2)^2 + (y + 3)^2 = 20$.

▶ Show solution

Centre $(2, -3)$. Radius $\sqrt{20} = 2\sqrt5$.

Question 6

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

▶ Show solution

$(x - 4)^2 - 16 + (y + 1)^2 - 1 - 8 = 0$

$(x - 4)^2 + (y + 1)^2 = 25$: centre $(4, -1)$, radius $5$.

Question 7

Does the point $(1, 2)$ lie inside, on or outside the circle in Question 6?

▶ Show solution

$(1 - 4)^2 + (2 + 1)^2 = 9 + 9 = 18$.

$18 < 25$, so the point is inside the circle.

Question 8

Find the values of $c$ for which $y = x + c$ is a tangent to $x^2 + y^2 = 18$.

▶ Show solution

$x^2 + (x + c)^2 = 18 \;\Rightarrow\; 2x^2 + 2cx + c^2 - 18 = 0$

Discriminant $= 4c^2 - 8(c^2 - 18) = 144 - 4c^2$

Tangent when $144 - 4c^2 = 0$, so $c^2 = 36$ and $c = \pm 6$.

Question 9

Find where the line $x + y = 7$ meets the circle $x^2 + y^2 = 25$.

▶ Show solution

$y = 7 - x$: $x^2 + (7 - x)^2 = 25 \;\Rightarrow\; 2x^2 - 14x + 24 = 0 \;\Rightarrow\; x^2 - 7x + 12 = 0$

$(x - 3)(x - 4) = 0$, so $x = 3$ or $4$.

Points $(3, 4)$ and $(4, 3)$.

Question 10

A garden is modelled on coordinate axes with units of metres. A sprinkler at $S(4, 3)$ waters every point within $5$ m of it. A straight path follows the line $y = x - 6$.

(a) Write down the equation of the boundary of the watered region. (b) Find the coordinates of the ends of the section of path that gets wet. (c) Find the length of path that gets wet, to 3 significant figures. (d) Find the shortest distance from the sprinkler to the path.

▶ Show solution

(a) $(x - 4)^2 + (y - 3)^2 = 25$

(b) Substitute $y = x - 6$: $(x - 4)^2 + (x - 9)^2 = 25$

$2x^2 - 26x + 97 - 25 = 0 \;\Rightarrow\; x^2 - 13x + 36 = 0 \;\Rightarrow\; (x - 4)(x - 9) = 0$

Ends at $(4, -2)$ and $(9, 3)$.

(c) $\sqrt{5^2 + 5^2} = 5\sqrt2 \approx 7.07$ m.

(d) The perpendicular from the centre bisects the chord, so the closest point is the midpoint $(6.5, 0.5)$.

Distance $= \sqrt{2.5^2 + 2.5^2} = 2.5\sqrt2 \approx 3.54$ m.