Printable · GCSE Higher · ages 14-16
Equation of a circle and its tangent worksheet — GCSE Higher
Fifteen questions on "equation of a circle and its tangent" — DfE statement A16. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
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Answer key: Equation of a circle and its tangent worksheet — GCSE Higher
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- (a) 12.5 — Method: the tangent at A is perpendicular to the radius OA, so find the gradient of OA, take its negative reciprocal, write the equation of the tangent and find where it meets the y-axis; the length of OB is then the distance of that crossing from the origin. Working: OA runs from (0, 0) to (−6, 8), so its gradient is 8 ÷ (−6), which cancels to −4/3; the negative reciprocal of −4/3 is 3/4. Substituting into y − 8 = 3/4(x + 6) gives y = 0.75x + 4.5 + 8, so y = 0.75x + 12.5 and B is (0, 12.5). The length OB is therefore 12.5. Answer: 12.5. The distractors: 10 is the radius of the circle, quoted on the assumption that the tangent always meets an axis one radius from the centre, which is only true when the radius itself lies along that axis; 8 is the y-coordinate of A, quoted by treating the tangent as horizontal so that it keeps the height of A; 3.5 comes from turning the gradient of OA upside down without changing its sign, which gives y = −0.75x + 3.5.
- (c) 14.1 — Method: in x² + y² = r² the right-hand side is the square of the radius, so take its square root to find the radius, then double the radius because the diameter is twice the radius. Working: r² = 50, so r = √50 = 7.07106…, and the diameter is 2 × 7.07106… = 14.14213…, which is 14.1 correct to 1 decimal place. Answer: 14.1. The distractors: 7.1 is the radius, worked out correctly but never doubled, so it answers a question about the radius rather than the diameter; 100.0 comes from doubling the 50 on the right-hand side, treating r² as though it were already the radius; 25.0 comes from halving the 50, treating r² as though it were already the diameter.
- (b) 9x + 40y = 1681 — For a circle x² + y² = r² centred at the origin, the tangent at a point (a, b) on the circle has equation ax + by = r². Here (a, b) = (9, 40) and r² = 1681, so the tangent is 9x + 40y = 1681. Choosing 40x + 9y = 1681 swaps the coefficients, using the y-coordinate as the x-coefficient and the x-coordinate as the y-coefficient. Choosing 9x + 40y = 41 uses the radius 41 instead of r² = 1681 as the constant. Choosing 9x − 40y = 1681 has the correct coefficients and constant but the wrong sign on the y-term.
- (b) 8 — Method: a circle centred on the origin has equation x² + y² = r², where r is the radius, so the number on the right-hand side is the square of the radius and not the radius itself. Working: comparing x² + y² = 64 with x² + y² = r² gives r² = 64, so r = √64 = 8. Answer: the radius is 8. The distractors: 64 is r² read straight off the equation as though the right-hand side were the radius, which is the commonest error on this form; 32 comes from halving 64, treating the right-hand side as a diameter that has to be halved; 16 is the diameter, 2 × 8, quoted in place of the radius.
- (c) 37.5 — The tangent at (18, 24) is 18x + 24y = 900 (using ax + by = r² with a = 18, b = 24, r² = 900). Setting x = 0 to find the y-intercept: 24y = 900, so y = 37.5. Choosing 900 skips the division by 24 and just repeats the constant. Choosing 50 divides the constant by the x-coefficient 18 instead of the y-coefficient 24. Choosing 1.25 uses the radius 30 instead of r² = 900 as the constant before dividing.
- (d) x = 4 and x = −4 — Substituting y = 3 gives x² + 9 = 25, which simplifies to x² = 16, so x = 4 or x = −4. Choosing 'x = 3 and x = −3' uses the given value y = 3 as if it were the x-coordinate. Choosing 'x = 4' alone finds the positive square root of 16 but forgets the negative root. Choosing 'x = 5 and x = −5' skips subtracting 3² = 9 from 25 and takes the square root of 25 directly.
- (a) 16 — A and B share the same y-coordinate, so the distance between them is the horizontal difference between their x-coordinates: 8 − (−8) = 16, giving a chord of length 16. Choosing 8 gives only the x-coordinate of one point, not the full distance between the two points. Choosing 34 is the diameter of the circle (2 × 17), not the length of this particular chord. Choosing 17 is the radius — the distance from the centre to A or to B, not from A to B.
- (d) −11/60 — The radius from the origin to (11, 60) has gradient 60/11. The tangent is perpendicular to this radius, so its gradient is the negative reciprocal: −1 ÷ (60/11) = −11/60. Choosing 60/11 uses the radius's gradient unchanged, without applying perpendicularity. Choosing −60/11 negates the radius's gradient but forgets to take its reciprocal. Choosing 11/60 takes the reciprocal correctly but keeps the gradient positive instead of negative.
- (a) y = (4/3)x + 25/3 — The radius from (0, 0) to (−4, 3) has gradient 3 ÷ (−4) = −3/4. The tangent is perpendicular to the radius, so its gradient is the negative reciprocal, 4/3. Using y − y₁ = m(x − x₁) with the point (−4, 3): y − 3 = (4/3)(x + 4), so y = (4/3)x + 16/3 + 3 = (4/3)x + 25/3. Using the radius's own gradient, −3/4, instead of taking the perpendicular gradient, gives y − 3 = (−3/4)(x + 4), which simplifies to y = −(3/4)x once the −3 and +3 in the constant cancel out. Taking the reciprocal of the radius's gradient but keeping the wrong sign, using −4/3 instead of 4/3, gives y = −(4/3)x − 7/3. Correctly finding the gradient 4/3 and expanding the bracket, but forgetting to add the y-coordinate 3 at the end, gives y = (4/3)x + 16/3.
- (b) x² + y² = 1369 — For a circle centred at the origin, the radius squared equals the sum of the squares of the coordinates of any point on it: r² = 12² + 35² = 144 + 1225 = 1369. The equation is x² + y² = 1369. x² + y² = 2209 comes from adding the coordinates first and then squaring the sum: (12 + 35)² = 47² = 2209, instead of squaring each coordinate separately. x² + y² = 1225 comes from using only 35² and leaving out the 12² term. x² + y² = 144 comes from using only 12² and leaving out the 35² term.
- (a) 842.0 — The radius to (20, 21) has gradient 21/20, so the tangent's gradient is −20/21. The tangent line is y − 21 = −20/21(x − 20), i.e. y = −20/21x + 841/21. Setting y = 0 gives the x-intercept x = 841/20 = 42.05; setting x = 0 gives the y-intercept y = 841/21 ≈ 40.048. The area of triangle OPQ is 1/2 × 42.05 × 40.048 ≈ 842.0. 1684.0 comes from multiplying the two intercepts without the 1/2 that a triangle's area needs — twice the correct area. 580.7 comes from using the circle's radius, 29, as a side of the triangle instead of the x-intercept, 42.05: 1/2 × 29 × 40.048 ≈ 580.7. 2.0 comes from a sign error in the tangent's gradient — using 20/21 instead of −20/21 — which gives a different line, with intercepts x ≈ −2.05 and y ≈ 1.952, and area 1/2 × 2.05 × 1.952 ≈ 2.0.
- (a) (15, 0) — Method: the tangent is perpendicular to the radius at the point of contact, so find the gradient of the radius, take its negative reciprocal, write the equation of the tangent, then substitute y = 0 because every point on the x-axis has y-coordinate 0. Working: the radius from (0, 0) to (3, 6) has gradient 6 ÷ 3 = 2, so the tangent has gradient −1/2. Substituting into y − 6 = −1/2(x − 3) gives y = −0.5x + 7.5. Setting y = 0 gives 0.5x = 7.5, so x = 15 and P is (15, 0). Answer: (15, 0). The distractors: (0, 7.5) is where the same tangent crosses the y-axis, reached by setting x = 0 instead of y = 0; (0, 0) comes from using the gradient of the radius, 2, for the tangent, which gives the line y = 2x through the centre and so crosses the x-axis at the origin; (6, 0) comes from changing the sign of the radius gradient without turning it upside down, which gives y = −2x + 12.
- (d) 2√2 — The radius satisfies r² = 8, so r = √8 = √(4 × 2) = √4 × √2 = 2√2. Choosing 8 forgets to take the square root of r² at all. Choosing 4 comes from halving 8 instead of finding its square root. Choosing √2 splits off the factor of 4 correctly but forgets to multiply the 2 back in front of the root.
- (b) x² + y² = 625 — Since (7, 24) lies on the circle, x² + y² = 7² + 24² = 49 + 576 = 625, so the equation is x² + y² = 625. Choosing x² + y² = 31 adds the coordinates 7 and 24 directly instead of squaring them first. Choosing x² + y² = 576 uses only 24² and forgets to add 7². Choosing x² + y² = 49 uses only 7² and forgets to add 24².
- (a) (0, 7) and (0, −7) — A tangent is horizontal where the radius to that point is vertical, i.e. where the point lies on the y-axis. On x² + y² = 49, setting x = 0 gives y² = 49, so y = 7 or y = −7. The points are (0, 7) and (0, −7). (7, 0) and (−7, 0) comes from swapping the condition — these are the points where the tangent is VERTICAL, not horizontal (the radius there is horizontal). (0, 7) only comes from finding one valid point but forgetting that y² = 49 also gives the negative root, y = −7. (7, 0) only combines both mistakes: the wrong axis, and only one of the two roots.
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