Practise › Questions › Circles
Circles questions
A circle is one sentence of algebra, every point at distance r from a fixed centre, and Pythagoras turns that sentence into an equation. Completing the square recovers centre and radius from a scrambled form, and three circle theorems from GCSE come back with coordinates and real jobs to do.
7 original questions · 22 marks · the circles notes · Coordinate geometry
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
Write down the equation of the circle with centre (2, −3) and radius 4.
Worked answer
(x − 2)2 + (y + 3)2 = 16. M1 for the form with the centre in place, A1 for the square of the radius. The signs flip on the way in, and the radius is squared. Both of the usual slips live in that one line.Write down the centre and radius of the circle (x + 1)2 + (y − 5)2 = 49.
Worked answer
Centre (−1, 5), radius √49 = 7. B1 for the centre, B1 for the radius. Reading (x + 1) as centre +1 is the flip in reverse. Quoting the radius as 49 is the other frequent miss, and it costs the second mark on its own.Find the centre and radius of the circle x2 + y2 − 4x + 6y − 3 = 0.
Worked answer
Complete the square in each variable. (x − 2)2 − 4 + (y + 3)2 − 9 − 3 = 0, so (x − 2)2 + (y + 3)2 = 16, giving centre (2, −3) and radius 4. M1 for completing the square in both variables, A1 for the centre, A1 for the radius. All three loose constants must migrate to the right before you take the square root; leaving the −3 behind is worth a mark. The same form settles where a point sits: substitute it into the left side and compare with 16, so (5, 1) gives 9 + 16 = 25 and lies outside.Find the equation of the tangent to the circle (x − 1)2 + (y − 2)2 = 25 at the point (4, 6), giving your answer in the form ax + by + c = 0.
Worked answer
Check the point first. 32 + 42 = 25, so it does lie on the circle. The radius from (1, 2) to (4, 6) has gradient 4/3, so the tangent's gradient is the negative reciprocal, −3/4. Through (4, 6): y − 6 = −(3/4)(x − 4), which clears to 3x + 4y − 36 = 0. M1 for the gradient of the radius, M1 for the negative reciprocal, A1 for the equation through the point, A1 for the demanded form. The perpendicular gradient and the demanded form are two separate pieces of work, so an answer left as y = … has finished only one of them. No calculus appears anywhere; the perpendicular-radius fact does all the work.A chord of a circle of radius 5 cm lies at perpendicular distance 3 cm from the centre. Find the length of the chord.
Worked answer
The perpendicular from the centre bisects the chord, so it makes a right triangle with hypotenuse 5 and one leg 3. The half-chord is √(25 − 9) = 4, and the chord is 8 cm. M1 for the right triangle with hypotenuse 5 and one leg 3, A1 for the chord. Stopping at 4 is the slip the word half exists to prevent.The line y = 2x − 3 meets the circle x2 + y2 − 6x + 4y − 12 = 0 at the points A and B. Find the coordinates of A and B.
Worked answer
Substitute the line into the circle so that only x survives: x2 + (2x − 3)2 − 6x + 4(2x − 3) − 12 = 0. Expanding gives 5x2 − 10x − 15 = 0, and dividing by 5 leaves x2 − 2x − 3 = 0, so (x − 3)(x + 1) = 0 and x = 3 or x = −1. Feeding each back into the line, A(3, 3) and B(−1, −5). M1 for substituting the line into the circle, A1 for the reduced quadratic, M1 for solving it, A1 for the x values, A1 for both points. Substitution must go into the circle, never the other way round, and the y values come from the line because the arithmetic there is trivial. Two real roots means the line is a secant; a repeated root would have meant a tangent, and no real root a clean miss.A(−5, 0) and B(5, 0) are the ends of a diameter of the circle x2 + y2 = 25, and C is the point (3, 4). Show that C lies on the circle, and verify that angle ACB is a right angle.
Worked answer
32 + 42 = 25, so C is on the circle. The vector from C to A is (−8, −4) and from C to B is (2, −4), giving gradients 1/2 and −2, whose product is −1. So CA is perpendicular to CB. B1 for the substitution showing C is on the circle, M1 for the two gradients, A1 for their values, A1 for the product and the conclusion. This is the angle-in-a-semicircle theorem caught in the act, and any other point of the circle would behave the same way. A show-that question wants the substitution written out; the theorem's name is no substitute for the working.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise circles one question at a time
The player marks nothing for you. It shows one question, waits, then shows the worked answer so you can mark yourself, and brings a question back sooner when it went badly.