Edexcel International A Level Maths · P2 · Pure Mathematics 2
Coordinate geometry of circles
28 free, high-quality past-paper-style questions on Coordinate geometry of circles for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 240 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P2-2-B-01P2 · ASEasy5 marks
The circle C has centre (4, −3) and radius 6.
(a)
Write down an equation for C.
(1)
(b)
Show that the equation of C can be written as x2 + y2 − 8x + 6y − 11 = 0.
(2)
(c)
Determine whether the point (8, 1) lies inside, on or outside C, showing your working.
The points A(−2, 5) and B(6, −1) are the ends of a diameter of the circle C.
Figure: the circle C with diameter AB and the point P
(a)
Find the coordinates of the centre of C.
(1)
(b)
Find the radius of C.
(2)
(c)
Write down an equation for C.
(1)
(d)
The point P(5, 6) lies on C. Show that angle APB is a right angle.
(2)
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Answer
Mark
Notes
Part (a)
midpoint of AB: (2, 2)
B1
Part (b)
AB2 = 82 + 62 = 100, or distance from the centre to A: √(16 + 9)
M1
radius = 5
A1
Part (c)
(x − 2)2 + (y − 2)2 = 25 (ft from (a) and (b))
B1
Part (d)
gradient AP = 1/7 and gradient BP = 7/(−1) = −7
B1
product = −1, so AP ⊥ BP and angle APB = 90° (alternatively: P is on the circle and AB is a diameter, so the angle in a semicircle is 90°, with P checked on C: 9 + 16 = 25)
B1
Exam tip: When a question says “show”, the reason must be written: give the gradient product −1, or name the property (angle in a semicircle) and check the point is on the circle.
The circle C has equation (x − 3)2 + (y + 1)2 = 25. The point P has coordinates (6, 3).
Figure: the circle C, the radius to P and the tangent at P
(a)
Verify that P lies on C.
(1)
(b)
Find the gradient of the radius from the centre of C to P.
(2)
(c)
Find an equation of the tangent to C at P, giving your answer in the form ax + by + c = 0, where a, b and c are integers.
(3)
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Answer
Mark
Notes
Part (a)
(6 − 3)2 + (3 + 1)2 = 9 + 16 = 25 ✓
B1
Part (b)
(3 − (−1)) ÷ (6 − 3)
M1
= 4/3
A1
Part (c)
tangent gradient = −3/4 (negative reciprocal)
M1
y − 3 = −¾(x − 6)
M1
3x + 4y − 30 = 0 (or any integer multiple)
A1
Exam tip: The tangent is perpendicular to the radius at the point of contact: find the gradient of the radius first, then take the negative reciprocal.
Figure 1 shows the circle C. The centre K of C and the point A, which lies on C, are both at grid points.
(a)
Use Figure 1 to find an equation for C in the form (x − p)2 + (y − q)2 = k.
(3)
(b)
The straight line through A and K meets C again at the point B. Find the coordinates of B.
(2)
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Answer
Mark
Notes
Part (a)
K(2, −1) and A(5, 1) read from the figure
B1
r2 = KA2 = 32 + 22 = 13
M1
(x − 2)2 + (y + 1)2 = 13
A1
Part (b)
AB is a diameter, so K is the midpoint of AB: B = (2 × 2 − 5, 2 × (−1) − 1)
M1
B(−1, −3)
A1
Common error: Writing (x + 2)2 + (y − 1)2. The signs inside the brackets are the opposite of the signs of the centre’s coordinates: centre (2, −1) gives (x − 2) and (y + 1).
The circle C, shown in Figure 1, has equation x2 + y2 − 8x + 10y + 16 = 0. C touches the x-axis at T and cuts the y-axis at A and B, where A is above B.
(a)
By completing the square, find the coordinates of the centre of C and the radius of C.
(3)
(b)
Explain why C touches the x-axis, and write down the coordinates of T.
(2)
(c)
Find the coordinates of A and B.
(2)
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Answer
Mark
Notes
Part (a)
(x − 4)2 − 16 + (y + 5)2 − 25 + 16 = 0
M1
centre (4, −5)
A1
(x − 4)2 + (y + 5)2 = 25, radius 5
A1
Part (b)
the centre is 5 units below the x-axis, which equals the radius, so the x-axis is a tangent
B1
T(4, 0)
B1
Part (c)
put x = 0: y2 + 10y + 16 = 0, so (y + 2)(y + 8) = 0
M1
A(0, −2) and B(0, −8)
A1
Exam tip: A circle touches an axis exactly when the distance from its centre to that axis equals the radius. Here |−5| = 5 for the x-axis, but |4| < 5 for the y-axis, so it cuts the y-axis twice.
Figure 1 shows a circle with centre C and a chord AB of the circle. The midpoint of AB is M. The points A, B and C lie on grid points.
(a)
Write down the coordinates of A and B, and find the coordinates of M.
(2)
(b)
Find an equation of the perpendicular bisector of AB, giving your answer in the form y = mx + c.
(3)
(c)
Verify that C lies on the line found in part (b), and find the exact length of CM.
(2)
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Answer
Mark
Notes
Part (a)
A(−6, 2) and B(1, 3)
B1
M(−2.5, 2.5)
B1
Part (b)
gradient of AB = 1/7
B1
perpendicular gradient −7: y − 2.5 = −7(x + 2.5)
M1
y = −7x − 15
A1
Part (c)
C(−2, −1): −7(−2) − 15 = −1 ✓
B1
CM = √(0.52 + 3.52) = √12.5 = 5√2/2
B1
Exam tip: The perpendicular from the centre to a chord bisects the chord, so the centre lies on the perpendicular bisector of every chord. Check: (5√2/2)2 + (½AB)2 = 12.5 + 12.5 = 25 = r2.
The circle C has equation x2 + y2 − 2x + 4y − 15 = 0. Figure 1 shows C together with three points D, E and F, each at a grid point.
(a)
Find the centre of C and the exact radius of C.
(2)
(b)
Determine, for each of the points D, E and F, whether it lies inside C, on C or outside C. You must show your working.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
(x − 1)2 + (y + 2)2 = 1 + 4 + 15
M1
centre (1, −2), radius √20 = 2√5
A1
Part (b)
D(5, 0), E(−2, 1), F(4, 2) read from the figure, and a squared distance from (1, −2) found, e.g. D: 42 + 22 = 20
M1
E: 32 + 32 = 18 and F: 32 + 42 = 25
A1
compared with 20: D on C, E inside, F outside
A1
Common error: Deciding from the picture alone. A point drawn close to the curve can be just inside or just outside, so compare each squared distance from the centre with r2 = 20.
Consider the circle C: x2 + y2 − 6x − 2y + 5 = 0, together with the family of straight lines through the origin l: y = mx, where m can take any real value.
(a)
Find the centre and the exact radius of C.
(2)
(b)
Show that the x-coordinates of any points where l meets C satisfy (1 + m2)x2 − (6 + 2m)x + 5 = 0.
(2)
(c)
Find the values of m for which l is a tangent to C.
(3)
(d)
Find the set of values of m for which l meets C at two distinct points.
(1)
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Answer
Mark
Notes
Part (a)
(x − 3)2 + (y − 1)2 = 9 + 1 − 5
M1
centre (3, 1), radius √5
A1
Part (b)
x2 + m2x2 − 6x − 2mx + 5 = 0
M1
(1 + m2)x2 − (6 + 2m)x + 5 = 0 (given answer)
A1
Part (c)
(6 + 2m)2 − 20(1 + m2) = 0
M1
−16m2 + 24m + 16 = 0, so 2m2 − 3m − 2 = 0
A1
(2m + 1)(m − 2) = 0: m = 2 or m = −½
A1
Part (d)
−½ < m < 2 (ft from (c))
B1
Model sketch
Figure: model sketch: the two tangents from the origin (m = 2 and m = −½)
Exam tip: The two tangent gradients multiply to −1 here, so the two tangents from the origin are perpendicular — a quick check that the values are plausible.
Exam tip: The centre is equidistant from all three points, so it lies on the perpendicular bisector of every chord. Check the radius using all three points.
The point A(1, 7) lies on the circle C whose equation is x2 + y2 + 4x − 6y − 12 = 0. The tangent to C at A forms a triangle with the two coordinate axes.
Figure: the circle C, the tangent at A and the triangle ORS
(a)
Find the centre and radius of C.
(3)
(b)
Find an equation of the tangent to C at A, giving your answer in the form ax + by + c = 0, where a, b and c are integers.
(3)
(c)
The tangent meets the x-axis at R and the y-axis at S. Find the exact area of triangle ORS, where O is the origin.
(2)
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Answer
Mark
Notes
Part (a)
(x + 2)2 + (y − 3)2 = 4 + 9 + 12
M1
centre (−2, 3)
A1
radius 5
A1
Part (b)
gradient of radius = (7 − 3)/(1 + 2) = 4/3
M1
tangent: y − 7 = −¾(x − 1)
M1
3x + 4y − 31 = 0
A1
Part (c)
R(31/3, 0) and S(0, 31/4)
M1
area = ½ × 31/3 × 31/4 = 961/24
A1
Common error: Using the gradient of the radius (4/3) as the gradient of the tangent. The tangent is perpendicular to the radius.
A circle lies in the first quadrant and touches both the x-axis and the y-axis. Its centre is (a, a), where a > 0. The circle passes through the point (2, 1).
(a)
Explain why the equation of the circle is (x − a)2 + (y − a)2 = a2.
(1)
(b)
Show that a2 − 6a + 5 = 0.
(3)
(c)
Hence find the equations of the two possible circles.
(2)
(d)
Find the exact distance between the centres of the two circles.
(2)
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Answer
Mark
Notes
Part (a)
the circle touches the x-axis, so the radius equals the distance from the centre to the axis, which is a
Figure 1 shows the plan of a rectangular lawn, with distances in metres. A sprinkler at the point S waters every point of the lawn within √10 m of S. A straight path crosses the lawn from U to V. The points S, U and V are at grid points.
(a)
Write down an equation of the boundary of the watered region.
(1)
(b)
Find an equation of the line through U and V.
(2)
(c)
Find the coordinates of the two points where the path crosses the boundary of the watered region.
(3)
(d)
Show that exactly half of the length of the path is watered.
Figure 1 shows triangle PQR, whose vertices are at grid points. The circle C through P, Q and R has equation x2 + y2 + ax + by + c = 0, where a, b and c are constants.
(a)
Read the coordinates of P, Q and R from Figure 1 and use them to form three equations in a, b and c.
(2)
(b)
Solve your equations to find a, b and c.
(3)
(c)
Find the centre of C and the exact radius of C.
(2)
(d)
Determine whether the origin lies inside or outside C.
(1)
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Answer
Mark
Notes
Part (a)
P(−3, 4), Q(3, 6), R(9, 0) substituted, e.g. 81 + 9a + c = 0
M1
−3a + 4b + c = −25, 3a + 6b + c = −45, 9a + c = −81
A1
Part (b)
eliminate c: 6a + 2b = −20 and 12a − 4b = −56
M1
a = −4, b = 2
A1
c = −45
A1
Part (c)
(x − 2)2 + (y + 1)2 = 4 + 1 + 45
M1
centre (2, −1), radius √50 = 5√2
A1
Part (d)
at (0, 0) the left-hand side is c = −45 < 0 (or distance √5 < 5√2), so the origin is insideC
B1
Exam tip: Each point gives a linear equation in a, b and c. Subtracting pairs of equations removes c at once and leaves two simple simultaneous equations.
The circle C has equation x2 + y2 − 4x − 10y + 20 = 0. Figure 1 shows C, its centre N, and the point P(k, 0) on the positive x-axis. The line PT touches C at T.
(a)
Find the coordinates of N and the radius of C.
(2)
(b)
Show that PT2 = k2 − 4k + 20.
(2)
(c)
Given that PT = 8, find the exact value of k.
(3)
(d)
Find the size of angle NPT, giving your answer in degrees to 1 decimal place.
Exam tip: Substituting P(k, 0) into the left-hand side of x2 + y2 − 4x − 10y + 20 gives k2 − 4k + 20 directly: for a point outside a circle, this value is the square of the tangent length.
Figure 1 shows the circle C with equation x2 + y2 + 4x − 2y − 3 = 0 and centre N. The circle meets the y-axis at A and B. The shaded region R is bounded by C and the y-axis and lies to the right of the y-axis.
(a)
Find the coordinates of N and the exact radius of C.
(2)
(b)
Find the coordinates of A and B.
(2)
(c)
Show that angle ANB = π/2 radians.
(2)
(d)
Find the exact area of R.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
(x + 2)2 + (y − 1)2 = 4 + 1 + 3
M1
N(−2, 1), radius √8 = 2√2
A1
Part (b)
x = 0: y2 − 2y − 3 = 0, so (y − 3)(y + 1) = 0
M1
A(0, 3) and B(0, −1)
A1
Part (c)
gradient NA = 2/2 = 1 and gradient NB = −2/2 = −1 (or NA2 + NB2 = 8 + 8 = 16 = AB2)
M1
product −1, so NA ⊥ NB and angle ANB = π/2
A1
Part (d)
sector − triangle: ½ × 8 × π/2 − ½ × 8
M1
2π − 4
A1
Common error: Using the area of the sector (2π) as the answer. The shaded region is a segment: subtract the area of triangle ANB from the area of the sector.
The circle C has equation x2 + y2 + 2x − 4y = 0 and centre N. Figure 1 shows C and the straight line ℓ, which crosses the coordinate axes at the two marked grid points.
(a)
Use Figure 1 to find an equation of ℓ.
(2)
(b)
Use algebra to show that ℓ does not meet C.
(3)
(c)
The line through N perpendicular to ℓ meets ℓ at F. Find the coordinates of F.
(2)
(d)
Hence find the shortest distance between a point on ℓ and a point on C, giving your answer to 3 significant figures.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
ℓ passes through (4, 0) and (0, −4)
B1
y = x − 4
B1
Part (b)
x2 + (x − 4)2 + 2x − 4(x − 4) = 0
M1
2x2 − 10x + 32 = 0, i.e. x2 − 5x + 16 = 0
A1
discriminant 25 − 64 = −39 < 0, so no real roots and no point of intersection
A1
Part (c)
N(−1, 2); perpendicular line y − 2 = −(x + 1)
M1
x − 4 = 1 − x: F(2.5, −1.5)
A1
Part (d)
NF = √(3.52 + 3.52) = 3.5√2 and radius √5
M1
3.5√2 − √5 = 2.71
A1
Exam tip: The shortest distance from a line to a circle lies along the perpendicular from the centre to the line: find the foot of the perpendicular, then subtract the radius.
In Figure 1, the points A(−1, 1) and B(7, 5) are the ends of a diameter of the circle C, whose centre is M. The point P lies on C, above AB, and has x-coordinate 5.
(a)
Find the coordinates of M.
(1)
(b)
Find an equation for C.
(2)
(c)
Find the y-coordinate of P.
(3)
(d)
Explain why angle APB = 90°, and hence find the area of triangle APB.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
M(3, 3)
B1
Part (b)
r2 = (3 + 1)2 + (3 − 1)2 = 20
M1
(x − 3)2 + (y − 3)2 = 20
A1
Part (c)
(5 − 3)2 + (y − 3)2 = 20, so (y − 3)2 = 16
M1
y = 7 or y = −1
A1
P is above AB (where AB has y = 4 at x = 5), so y = 7
A1
Part (d)
AB is a diameter, so angle APB is an angle in a semicircle and equals 90°
B1
AP = 6√2, BP = 2√2: area = ½ × 6√2 × 2√2 = 12
B1
Exam tip: A quadratic in y usually gives two points on the circle with the same x-coordinate; use the information in the figure (P is above AB) to choose the right one, and say why.
The circle C has equation x2 + y2 − 2x − 4y = 0. The point P has coordinates (6, 2). Two tangents are drawn from P to C, touching C at the points A and B.
Figure: the circle C and the two tangents from P (not to scale)
(a)
Find the centre and the exact radius of C.
(2)
(b)
A line through P has gradient m. Show that the x-coordinates of any points where this line meets C satisfy (1 + m2)x2 − (2 + 12m2)x + 36m2 − 4 = 0.
(3)
(c)
Find the gradients of the two tangents.
(3)
(d)
Find the coordinates of A and B.
(3)
(e)
Find the exact length of PA and the area of the quadrilateral PADB, where D is the centre of C.
PD = 5, so PA2 = 25 − 5 (or directly from P and A)
M1
PA = 2√5
A1
area = 2 × ½ × √5 × 2√5 = 10
A1
Exam tip: The configuration is symmetric about the line through P and the centre (here y = 2), so the second point of contact is the reflection of the first — a quick check on your algebra.
Figure: model sketch: C with centre (5, 4) and radius 5, and the tangent touching at (8, 8)
Exam tip: The nearest point of a circle to an outside point lies on the line through the centre, so the shortest distance is (distance to the centre) − radius.
Figure 1 shows the circles C1 and C2. C1 has equation x2 + y2 + 6x − 4y − 7 = 0 and centre A; C2 has equation (x − 3)2 + (y − 5)2 = 5 and centre B. The circles touch at the point T, and their common tangent at T crosses the x-axis at D.
(a)
Find the coordinates of A and the exact radius of C1.
(2)
(b)
Show that the circles touch.
(3)
(c)
Find the coordinates of T.
(2)
(d)
Find an equation of the common tangent at T.
(2)
(e)
Find the coordinates of D and the exact length of DT.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
(x + 3)2 + (y − 2)2 = 9 + 4 + 7
M1
A(−3, 2), radius √20 = 2√5
A1
Part (b)
B(3, 5) and radius of C2 = √5
B1
AB = √(62 + 32) = √45 = 3√5
M1
2√5 + √5 = 3√5 = AB, so the circles touch (externally)
A1
Part (c)
T lies on AB with AT : TB = 2√5 : √5 = 2 : 1, so T = (−3 + ⅔ × 6, 2 + ⅔ × 3)
M1
T(1, 4)
A1
Part (d)
gradient of AB = ½, so the tangent has gradient −2: y − 4 = −2(x − 1)
M1
y = 6 − 2x (or 2x + y − 6 = 0)
A1
Part (e)
D(3, 0)
B1
DT = √(22 + 42) = 2√5
B1
Exam tip: Two circles touch externally when the distance between the centres equals the sum of the radii, and internally when it equals the difference. The point of contact lies on the line of centres.
Figure 1 shows a map on a grid with distances in kilometres. A radar station at L detects any ship within √65 km of L. A ship sails in a straight line from S, through W, at a constant speed of 18 km h−1. The points L, S and W are at grid points.
(a)
Write down an equation of the boundary of the region in which ships are detected.
(1)
(b)
Find an equation of the ship’s course.
(2)
(c)
Use algebra to find the coordinates of the point E where the ship enters the detection region and the point Z where it leaves.
(4)
(d)
Find, in minutes to 3 significant figures, the time for which the ship is detected.
(2)
(e)
Find the shortest distance between the ship and L during the voyage, giving your answer in km to 3 significant figures.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
(x − 6)2 + (y − 5)2 = 65
B1
Part (b)
S(−5, −3) and W(0, 2): gradient = 5/5 = 1
M1
y = x + 2
A1
Part (c)
(x − 6)2 + (x − 3)2 = 65
M1
2x2 − 18x − 20 = 0, i.e. x2 − 9x − 10 = 0
A1
x = −1 or x = 10
A1
E(−1, 1) and Z(10, 12)
A1
Part (d)
EZ = √(112 + 112) = 11√2 km
M1
11√2 ÷ 18 h = 51.9 minutes
A1
Part (e)
the closest point is the foot of the perpendicular from L, which is the midpoint of the chord EZ: (4.5, 6.5)
M1
distance2 = 1.52 + 1.52 = 4.5
M1
√4.5 = 2.12 km
A1
Exam tip: The perpendicular from the centre of a circle to a chord bisects the chord, so the point of the course nearest to L is simply the midpoint of EZ.
Figure 1 is a sketch of the circle C with centre K. The line ℓ, with equation x + y = 9, touches C at the point P(6, 3). The circle also passes through the point Q(−6, −1). The tangent to C at Q meets ℓ at R.
(a)
Find an equation of the line through P that is perpendicular to ℓ.
(2)
(b)
Find an equation of the perpendicular bisector of PQ.
(3)
(c)
Hence find the coordinates of K and an equation for C.
(3)
(d)
Find an equation of the tangent to C at Q.
(2)
(e)
Find the coordinates of R and show that RP = RQ.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
gradient of ℓ = −1, so the perpendicular gradient is 1
M1
y = x − 3
A1
Part (b)
midpoint of PQ = (0, 1)
B1
gradient PQ = 4/12 = ⅓, so the perpendicular gradient is −3
M1
y = 1 − 3x
A1
Part (c)
K lies on both lines: x − 3 = 1 − 3x
M1
K(1, −2)
A1
r2 = 52 + 52 = 50: (x − 1)2 + (y + 2)2 = 50
A1
Part (d)
gradient KQ = 1/(−7) = −1/7, so the tangent has gradient 7
M1
y = 7x + 41
A1
Part (e)
7x + 41 = 9 − x, so x = −4
M1
R(−4, 13)
A1
RP2 = 102 + 102 = 200 and RQ2 = 22 + 142 = 200, so RP = RQ = 10√2
A1
Exam tip: The centre lies on the normal at a point of contact and on the perpendicular bisector of any chord. Two such lines are enough to locate it.
Figure 1 shows the circle C with equation x2 + y2 − 2x + 6y − 15 = 0 and centre K, and the point T(7, 5). The two tangents from T touch C at A and B. The shaded region R is bounded by TA, TB and the minor arc AB.
(a)
Find the coordinates of K and the radius of C.
(2)
(b)
Find the length KT, and show that TA = 5√3.
(3)
(c)
Show that angle AKB = 2π/3 radians.
(2)
(d)
Find the exact area of R.
(3)
(e)
Find the shortest distance from T to a point of C, and the coordinates of the point of C nearest to T.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
(x − 1)2 + (y + 3)2 = 1 + 9 + 15
M1
K(1, −3), radius 5
A1
Part (b)
KT2 = 62 + 82 = 100, KT = 10
B1
angle KAT = 90°, so TA2 = 100 − 25
M1
TA = √75 = 5√3 (given answer)
A1
Part (c)
cos AKT = 5/10 = ½, so angle AKT = π/3
M1
by symmetry about KT, angle AKB = 2 × π/3 = 2π/3
A1
Part (d)
area of kite KATB = 2 × ½ × 5 × 5√3 = 25√3
M1
area of sector KAB = ½ × 52 × 2π/3 = 25π/3
M1
area of R = 25√3 − 25π/3
A1
Part (e)
KT − radius = 10 − 5 = 5
B1
the nearest point is on KT, 5 from K: the midpoint of KT, (4, 1)
B1
Common error: Subtracting the whole sector from the triangle KAT only. Region R is the kite KATB (two right-angled triangles) minus the sector with angle 2π/3.
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