S5 Mathematics Paper: Equations Of Circles And Circle Properties (100 Marks)

S5 Mathematics Paper: Equations of Circles and Circle Properties (100 marks)

2026-11-11

S5 Mathematics – circle equations

S5 Mathematics paper on circle equations, tangency, radical axis and circle theorems, with full answers and marking notes.

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Paper details　S5 Mathematics practice paper with six questions covering equations of circles, distance and tangency conditions, the tangent at a point on a circle, the relative position of two circles and the radical axis, circle theorems and chord length. Full worked answers with marking notes.

Section A: Equations of Circles, Centres and Radii (16 marks)

The general equation of a circle is x squared plus y squared plus Dx plus Ey plus F equals zero. Completing the square turns it into the standard form, from which the centre and radius can be read off. To place a point relative to a circle, compare its distance from the centre with the radius.

#	Question	Answer and marking notes
1	Circle C has equation x^2 + y^2 – 6x + 8y – 24 = 0. Use completing the square to find its centre and radius.	Answer: completing the square gives (x – 3)^2 + (y + 4)^2 = 49, so the centre is (3, -4) and the radius is 7. (Marking: 1 mark per completed square, 1 mark for the constant, 0.5 mark each for centre and radius.)
2	State the position of the point P(3, 3) relative to circle C, with a reason.	Answer: the distance from P to the centre is 7 minus negative 4, that is 7, which equals the radius, so P lies on the circle. (Marking: 2 marks for the distance, 1 mark comparison, 1 mark conclusion.)
3	State the position of the point Q(9, -4) relative to circle C, with a reason.	Answer: the distance from Q to the centre is 6, which is less than the radius 7, so Q lies inside the circle. (Marking: 2 marks for the distance, 1 mark comparison, 1 mark conclusion.)
4	Find the number of intersections of circle C with the y-axis, with a reason.	Answer: putting x equal to zero gives y^2 + 8y – 24 = 0. The discriminant is 64 plus 96, that is 160, which is positive, so there are 2 intersections. (Marking: 1 mark for the quadratic, 2 marks discriminant, 1 mark conclusion.)

Section B: Tangency of a Line and a Circle (18 marks)

A line is tangent to the circle x squared plus y squared equals r squared when the distance from the centre to the line equals the radius. The point of contact is the foot of the perpendicular from the centre to the line.

#	Question	Answer and marking notes
5	Line L has equation y = x + k and is tangent to the circle x^2 + y^2 = 25. Find the two possible values of k.	Answer: rewrite the line as x – y + k = 0. The distance from the centre is the modulus of k divided by root 2, which equals 5, so k is plus or minus 5 root 2, about 7.1 and -7.1. (Marking: 2 marks distance formula, 2 marks equate to radius, 1 mark per value.)
6	Taking the positive value of k, find the coordinates of the point of contact, in surd form and as a decimal approximation.	Answer: the foot of the perpendicular is at minus k over 2 and k over 2, that is minus 5 root 2 over 2 and 5 root 2 over 2, about minus 3.5 and 3.5. (Marking: 2 marks for the foot, 1 mark per coordinate, 2 marks approximation.)
7	Taking the positive value of k, find the area of the triangle formed by L and the two coordinate axes.	Answer: the intercepts are at (0, k) and (-k, 0), both legs have length k, so the area is k squared over 2, that is 50 over 2, equal to 25 square units. (Marking: 1 mark per intercept, 2 marks formula, 2 marks answer.)

Section C: The Tangent at a Point on a Circle (16 marks)

The radius drawn to the point of contact is perpendicular to the tangent, so the product of their gradients is −1. First find the gradient of the radius, take the negative reciprocal, then write the equation in point-slope form.

#	Question	Answer and marking notes
8	Circle C has equation x^2 + y^2 – 4x – 6y – 12 = 0. Find its centre and radius.	Answer: the equation becomes (x – 2)^2 + (y – 3)^2 = 25, so the centre is (2, 3) and the radius is 5. (Marking: 2 marks completing the square, 1 mark centre, 1 mark radius.)
9	Show that the point A(5, 7) lies on circle C.	Answer: the square of the distance from A to the centre is 3 squared plus 4 squared, that is 25, which equals the square of the radius, so A lies on the circle. (Marking: 2 marks substitution, 1 mark comparison, 1 mark conclusion.)
10	Find the equation of the tangent to circle C at A, in general form.	Answer: the gradient of the radius is 4 over 3, so the tangent has gradient minus 3 over 4. Then y minus 7 equals minus 3 over 4 times x minus 5, which simplifies to 3x + 4y – 43 = 0. (Marking: 1 mark radius gradient, 1 mark tangent gradient, 2 marks equation.)
11	Find the x-intercept of that tangent.	Answer: putting y equal to zero gives 3x equal to 43, so the x-intercept is about 14.3. (Marking: 2 marks substitution, 1 mark solving, 1 mark approximation.)

Section D: Relative Position of Two Circles and the Radical Axis (18 marks)

Comparing the distance between centres with the sum and the difference of the radii decides whether two circles meet, touch externally, touch internally or are separate. Subtracting the two equations cancels the quadratic terms and leaves the radical axis.

#	Question	Answer and marking notes
12	Circle C1 is x^2 + y^2 – 4x = 0 and circle C2 is x^2 + y^2 – 6y + 8 = 0. Find the centre and radius of each circle.	Answer: C1 has centre (2, 0) and radius 2; C2 has centre (0, 3) and radius 1. (Marking: 1 mark per centre, 1 mark per radius.)
13	Determine the relative position of the two circles, with a reason.	Answer: the distance between centres is root 13, about 3.6, which is greater than the sum of radii 3, so the circles are separate and do not meet. (Marking: 2 marks distance, 1 mark comparison, 1 mark conclusion.)
14	Subtract the two equations to find the radical axis, in general form.	Answer: subtracting gives minus 4x plus 6y minus 8 equal to zero, that is 2x – 3y + 4 = 0. (Marking: 2 marks subtraction, 2 marks constants, 2 marks simplification.)
15	Decide whether the point (2, 3) lies on the radical axis, with a reason.	Answer: substitution gives 4 minus 9 plus 4, that is -3, which is not zero, so the point is not on the radical axis. (Marking: 2 marks substitution, 1 mark comparison, 1 mark conclusion.)

Section E: Circle Theorems (16 marks)

Opposite angles of a cyclic quadrilateral are supplementary, angles in the same segment are equal, and the angle between a tangent and a chord equals the angle in the alternate segment. These three results carry most senior secondary circle questions.

#	Question	Answer and marking notes
16	In cyclic quadrilateral ABCD, angle A is 3x plus 10 degrees and angle C is 2x plus 30 degrees. Find x, angle A and angle C.	Answer: opposite angles add to 180 degrees, so 3x plus 10 plus 2x plus 30 equals 180, giving x equal to 28, angle A equal to 94 degrees and angle C equal to 86 degrees. (Marking: 2 marks equation, 2 marks solving, 1 mark per angle.)
17	A, B, C and D lie on the same circle, with angle BAC equal to 34 degrees and angle ACB equal to 58 degrees. Find angle ABC and angle BDC.	Answer: the angles of triangle ABC add to 180 degrees, so angle ABC is 88 degrees; angles in the same segment are equal, so angle BDC equals angle BAC, that is 34 degrees. (Marking: 2 marks for angle ABC, 1 mark for the theorem, 2 marks for angle BDC.)
18	The tangent at A makes an angle of 52 degrees with chord AB. Find angle ACB, the angle in the alternate segment.	Answer: by the tangent-chord angle theorem, angle ACB equals 52 degrees. (Marking: 3 marks for naming the theorem, 2 marks answer with unit.)

Section F: Intersection of a Line and a Circle, Chord Length and Midpoint (16 marks)

Substituting the line into the circle gives a quadratic whose discriminant fixes the number of intersections. The midpoint of a chord is the foot of the perpendicular from the centre, so it can be found without solving for the two intersection points.

#	Question	Answer and marking notes
19	Circle C is x^2 + y^2 – 2x + 4y – 20 = 0. Find its centre and radius.	Answer: the equation becomes (x – 1)^2 + (y + 2)^2 = 25, so the centre is (1, -2) and the radius is 5. (Marking: 2 marks completing the square, 1 mark centre, 1 mark radius.)
20	The line y = x + 1 meets circle C. Use substitution to show that there are two intersections.	Answer: substitution gives 2x squared plus 4x minus 15 equal to zero. The discriminant is 16 plus 120, that is 136, which is positive, so there are two intersections. (Marking: 2 marks substitution, 1 mark discriminant, 1 mark conclusion.)
21	Find the length of the chord joining the two intersections, in surd form and as a decimal approximation.	Answer: the difference of the roots is root 136 over 2, that is root 34, so the chord is root 2 times root 34, equal to 2 root 17, about 8.2. (Marking: 2 marks root difference, 1 mark factor root 2, 0.5 mark each for surd and decimal.)
22	Find the coordinates of the midpoint of the chord and check them by the perpendicular-foot method.	Answer: the midpoint has x equal to minus 4 over 4, that is -1, and y equal to 0, so it is (-1, 0); the perpendicular foot is also (-1, 0), so both methods agree. (Marking: 2 marks midpoint, 1 mark check, 1 mark conclusion.)

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