Questions on circles

Question 1

Write down the equation of the circle with centre (2, −3) and radius 4.

Accepted Answer

(x − 2)² + (y + 3)² = 16. M1 for the form with the centre in place, A1 for the square of the radius. Each coordinate of the centre changes sign inside its bracket, and the right-hand side is the square of the radius.

Question 2

Write down the centre and radius of the circle (x + 1)² + (y − 5)² = 49.

Accepted 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.

Question 3

The line y = x meets the circle x² + y² − 4x + 2 = 0. Substitute the line into the circle to obtain a quadratic in x, and use its discriminant to state how many points the line and the circle have in common.

Accepted Answer

Substituting y = x gives x² + x² − 4x + 2 = 0, that is 2x² − 4x + 2 = 0, or x² − 2x + 1 = 0. M1 for the substitution and the quadratic. The discriminant is (−2)² − 4(1)(1) = 0. A1 for the discriminant. A zero discriminant means a repeated root, so the line touches the circle at exactly one point: it is a tangent. B1 for the conclusion. The repeated root itself is x = 1, giving the point of contact (1, 1), which does lie on the circle, 1 + 1 − 4 + 2 = 0.

Question 4

Circle C₁ has centre (0, 0) and radius 3. Circle C₂ has centre (6, 8) and radius 7. Find the distance between the centres, and state, with a reason, how the two circles lie.

Accepted Answer

d = √(6² + 8²) = √100 = 10. M1 for the distance, A1 for d = 10. The sum of the radii is 3 + 7 = 10, equal to d, so the circles touch externally, meeting at a single point on the line joining the centres. B1 for the comparison and the conclusion.

Question 5

Find the centre and radius of the circle x² + y² − 4x + 6y − 3 = 0.

Accepted 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 gives the wrong radius. 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.

Question 6

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.

Accepted 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.

Question 7

A chord of a circle of radius 5 cm lies at perpendicular distance 3 cm from the centre. Find the length of the chord.

Accepted 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 gives only the half-chord.

Question 8

A(1, 2), B(9, 8) and C(1, 8) lie on a circle. Show that angle ACB is a right angle, state what this makes AB, and hence find the equation of the circle through A, B and C.

Accepted Answer

The vector from C to A is (1 − 1, 2 − 8) = (0, −6) and from C to B is (9 − 1, 8 − 8) = (8, 0). Their dot product is 0×8 + (−6)×0 = 0, so CA is perpendicular to CB and angle ACB = 90°. M1 for the two vectors, A1 for the dot product and the right angle. By the converse of the angle in a semicircle, a right angle at C means AB must be a diameter of the circle through A, B and C. B1 for the converse, correctly applied. The midpoint of AB is (5, 5), and the radius is half of |AB| = ½√(8² + 6²) = ½ × 10 = 5. M1 for the midpoint and the radius. The circle is (x − 5)² + (y − 5)² = 25. A1 for the equation. Substituting C into it, (1 − 5)² + (8 − 5)² = 16 + 9 = 25, which checks that C lies on this circle.

Question 9

The circle x² + y² = 8 has a tangent of the form y = x + k. Substitute the line into the circle to form a quadratic in x, and use the discriminant to find the two possible values of k.

Accepted Answer

Substituting, x² + (x + k)² = 8, which expands to 2x² + 2kx + (k² − 8) = 0. M1 for the substitution, A1 for the quadratic. A tangent meets the circle at a single repeated point, so the discriminant is zero: (2k)² − 4(2)(k² − 8) = 0, that is 4k² − 8k² + 64 = 0, so k² = 16 and k = 4 or k = −4. M1 for setting the discriminant to zero, A1 for both values. The two tangents sit symmetrically on either side of the circle, which is why the two values of k are equal in size and opposite in sign.

Question 10

Circle C₁ has centre (0, 0) and radius 6. Circle C₂ has centre (4, 3) and radius 2. By comparing d² with (r₁ + r₂)² and (r₁ − r₂)², where d is the distance between the centres, decide how C₁ and C₂ lie, without evaluating any square root.

Accepted Answer

d² = 4² + 3² = 25. (r₁ + r₂)² = 8² = 64 and (r₁ − r₂)² = 4² = 16. M1 for d², A1 for the two squared bounds. Since 16 < 25 < 64, the circles cross at two points. B1 for the comparison and the conclusion. Every quantity compared is already a length squared, so no square root is needed at any stage. B1 for identifying why the method avoids one.

Question 11

The circle has equation (x − 2)² + (y + 1)² = 9, and P is the point (6, 2), outside the circle. Find the length of the tangent from P to the circle.

Accepted Answer

The centre is (2, −1) and the radius is 3. The distance from P to the centre is √((6 − 2)² + (2 − (−1))²) = √(16 + 9) = √25 = 5. M1 for the distance to the centre, A1 for 5. The tangent, the radius to its point of contact, and the line from the centre to P form a right triangle with that distance as hypotenuse, so the tangent length is √(5² − 3²) = √16 = 4. M1 for the right triangle with hypotenuse 5 and one leg 3, A1 for the tangent length.

Question 12

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.

Accepted 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.

Question 13

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.

Accepted 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. A show-that question requires the substitution to be written out; the theorem's name is no substitute for the working.

Question 14

The circles x² + y² − 6x − 2y − 15 = 0 and x² + y² − 18x − 2y + 57 = 0 are given. Show that they cross at two points, and find the coordinates of both.

Accepted Answer

Completing the square in each: (x − 3)² + (y − 1)² = 25 and (x − 9)² + (y − 1)² = 25, so the centres are (3, 1) and (9, 1), each with radius 5. M1 for completing the square in both, A1 for both centres and radii. The centres share a y-coordinate, so d = 9 − 3 = 6. The sum of the radii is 10 and the difference is 0, and 0 < 6 < 10, so the circles cross at two points. M1 for the comparison, A1 for the conclusion. Subtracting the second equation from the first cancels the squared terms and the y terms: 12x − 72 = 0, so x = 6, the common chord. M1 for the subtraction. Substituting into the first circle, (6 − 3)² + (y − 1)² = 25, so (y − 1)² = 16 and y = 5 or y = −3. The crossings are (6, 5) and (6, −3). A1 for both points. Both satisfy the second equation too, (6 − 9)² + (5 − 1)² = 9 + 16 = 25 and (6 − 9)² + (−3 − 1)² = 9 + 16 = 25, which checks them.

Question 15

A(0, 0), B(8, 0) and C(1, 7) are three points. Show that no angle of triangle ABC is a right angle, and hence find the equation of the circle through A, B and C using the perpendicular bisectors of two of its sides.

Accepted Answer

AB = (8, 0) and AC = (1, 7): their dot product is 8, not zero, so the angle at A is not 90°. BA = (−8, 0) and BC = (−7, 7): dot product 56, not zero, so B is not 90° either. CA = (−1, −7) and CB = (7, −7): dot product −7 + 49 = 42, not zero, so neither is C. B1 for checking all three and concluding no shortcut applies. The perpendicular bisector of AB has midpoint (4, 0), and since AB is horizontal the bisector is the vertical line x = 4. M1 for this bisector. The perpendicular bisector of AC has midpoint (0.5, 3.5); AC has gradient 7, so the bisector has gradient −1/7, giving y − 3.5 = −(1/7)(x − 0.5). Substituting x = 4: y = 3.5 − (1/7)(3.5) = 3.5 − 0.5 = 3. M1 for the second bisector and its equation, A1 for the centre (4, 3). The radius is the distance to A: √(4² + 3²) = 5, so the circle is (x − 4)² + (y − 3)² = 25. A1 for the radius and the equation. This agrees with B and C: (8 − 4)² + 3² = 25 and (1 − 4)² + (7 − 3)² = 25.