This mini-lesson covers AQA section C β Coordinate geometry in the (x, y) plane: straight lines and their gradients, parallel and perpendicular conditions, the equation of a circle and finding its centre and radius by completing the square, the three circle properties the spec names, and parametric equations.
Work through each screen, answer the questions as you go (a few are conceptual, most are calculations) and collect β stars. Everything here is A-level standard, drawn from the AQA 7357 subject content. Press Start when you are ready.
Coordinate geometry Β· lines
Straight lines: the three forms
Three equivalent ways of writing a line β each is convenient in a different situation.
y = mx + c β gradient m, y-intercept c. Best for reading off a gradient.
y β yβ = m(x β xβ) β the workhorse when you know a point and a gradient.
ax + by + c = 0 β the form AQA often demands for a final answer (integer coefficients).
m = (yβ β yβ)/(xβ β xβ)rise over run β subtract in a consistent order
Worked example β line through two points
Through A(2, 5) and B(6, 13): m = (13 β 5)/(6 β 2) = 8/4 = 2.
Using A: y β 5 = 2(x β 2) β y = 2x + 1 β in the required form, 2x β y + 1 = 0.
Coordinate geometry Β· lines
Parallel and perpendicular
Two straight lines with gradients mβ and mβ:
Parallel β mβ = mβ.
Perpendicular β mβ mβ = β1, i.e. mβ is the negative reciprocal of mβ.
Do not forget to subtract the 9 and the 16 that completing the square introduces β and remember to take the square root at the end.
Sort it
Parallel, perpendicular or neither?
Tap a pair of lines, then tap the relationship between them. Rearrange to y = mx + c first!
β₯ Parallel
β₯ Perpendicular
βοΈ Neither
Coordinate geometry Β· circles
The three circle properties
AQA names three circle facts, and every circle question leans on one of them:
The angle in a semicircle is a right angle β so if β APB = 90Β°, then AB is a diameter, and the centre is the midpoint of AB.
The perpendicular from the centre to a chord bisects the chord β so the perpendicular bisector of any chord passes through the centre.
The tangent is perpendicular to the radius at the point of contact.
Worked example β a tangent
Circle centre C(3, β4), radius 6. Find the tangent at P(3, 2) β a point on the circle, since (3β3)Β² + (2+4)Β² = 36 β
Radius CP is vertical (both x-coordinates are 3), so the tangent is horizontal: y = 2.
In general: find the gradient of the radius, take its negative reciprocal, then use y β yβ = m(x β xβ).
Inside, on or outside? Compare the distance from the centre with r. Distance from (3, β4) to the origin is β(3Β² + 4Β²) = 5, and 5 < 6, so the origin lies inside this circle.
Quick check
Tangent and radius
?What is the relationship between a tangent to a circle and the radius drawn to the point of contact?
Coordinate geometry Β· circles
Intersections: line and circle
To find where a line meets a circle, substitute the line into the circle equation. You get a quadratic in x, and its discriminant answers the geometry:
Ξ > 0 β the line is a secant: two intersection points.
Ξ = 0 β the line is a tangent: one repeated point of contact.
Ξ < 0 β the line misses the circle entirely.
tangency β discriminant = 0this is how "find k such that the line is a tangent" questions are done
Alternative for tangency: the perpendicular distance from the centre to the line equals r. Either method is acceptable β the discriminant route is usually safer in an exam because it needs no formula sheet.
Calculate
Your turn β a gradient
1Find the gradient of the straight line through A(2, 5) and B(6, 13).
2Find the gradient of any line perpendicular to 3x + 4y = 12. Give your answer as a decimal to 2 d.p.
Hint: Rearrange first: 4y = β3x + 12 β y = βΒΎx + 3, so mβ = β3/4. The perpendicular gradient is the negative reciprocal, 4/3 = 1.33 (2 d.p.).
Coordinate geometry Β· parametric
Parametric equations
Instead of linking x and y directly, a parameter t drives both. Each value of t gives one point on the curve.
x = f(t), y = g(t)eliminate t to get the Cartesian equation
Worked example β eliminating the parameter
x = 3t, y = tΒ² + 1. From the first equation, t = x/3.
Substitute: y = (x/3)Β² + 1 = xΒ²/9 + 1 β a parabola.
So when x = 12, t = 4 and y = 4Β² + 1 = 17 (and the Cartesian form agrees: 144/9 + 1 = 16 + 1 = 17 β).
Domain restrictions survive. If t β₯ 0 then x = 3t means x β₯ 0 β the Cartesian equation alone would wrongly suggest the whole parabola.
Quick check
Perpendicular line
?Find the equation of the line through (1, 2) that is perpendicular to y = 2x + 3.
Calculate
Your turn β radius of a circle
3Find the radius of the circle xΒ² + yΒ² β 6x + 8y β 11 = 0.
Hint: Complete the square: (x β 3)Β² β 9 + (y + 4)Β² β 16 β 11 = 0 β (x β 3)Β² + (y + 4)Β² = 36. The radius is β36, not 36.
Coordinate geometry Β· parametric
Trigonometric parameters
When x and y involve sin and cos, eliminate the parameter using the Pythagorean identity rather than by rearranging.
Worked example β a circle in disguise
x = cos t, y = sin t. Then xΒ² + yΒ² = cosΒ²t + sinΒ²t = 1 β the unit circle.
More generally x = a + r cos t, y = b + r sin t traces the circle centre (a, b), radius r, because (x β a)Β² + (y β b)Β² = rΒ²(cosΒ²t + sinΒ²t) = rΒ².
Sketch it. A rough diagram catches sign errors and tells you instantly whether an answer is plausible.
Quick check
Eliminate the parameter
?A curve has parametric equations x = cos t, y = sin t. What is its Cartesian equation?
Calculate
Your turn β inside or outside?
4That circle has centre (3, β4). Find the distance from the centre to the origin (0, 0). (Compare it with the radius 6 to decide whether the origin lies inside the circle.)
Hint: Distance = β((3 β 0)Β² + (β4 β 0)Β²) = β(9 + 16) = β25. Since 5 < 6, the origin lies inside the circle.
Calculate
Your turn β parametric point
5A curve has parametric equations x = 3t and y = tΒ² + 1. Find the y-coordinate of the point where x = 12.
Hint: x = 12 β 3t = 12 β t = 4. Then y = 4Β² + 1 = 17. (Cartesian check: y = xΒ²/9 + 1 = 144/9 + 1 = 17 β)