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.
Higher only
Answer key: Equation of a circle and its tangent worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- (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.
- (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.
- (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) (0, 6) and (0, −6) — Method: every point on the y-axis has x-coordinate 0, so substitute x = 0 into the equation of the circle and solve for y, remembering that a square root has a negative value as well as a positive one. Working: putting x = 0 into x² + y² = 36 leaves y² = 36, so y = 6 or y = −6, and the two crossings are (0, 6) and (0, −6). Answer: (0, 6) and (0, −6). The distractors: (0, 36) and (0, −36) use 36 itself as the distance from the centre, which reads r² as r; (6, 0) and (−6, 0) are the right distance from the centre but are the crossings of the x-axis, found by setting y = 0 instead of x = 0; (0, 18) and (0, −18) halve 36, treating the right-hand side of the equation as a diameter.
- (b) x² + y² = 169 — Method: a circle centred on the origin has equation x² + y² = r², and every point on it satisfies that equation, so substituting the coordinates of a point that lies on the circle gives r² directly. Working: substituting x = 5 and y = 12 gives 5² + 12² = 25 + 144 = 169, so r² = 169 and the circle is x² + y² = 169. Answer: x² + y² = 169. The distractors: x² + y² = 13 uses the radius, √169 = 13, where r² belongs, which is the confusion between r and r² made in the other direction; x² + y² = 17 adds the two coordinates, 5 + 12, instead of adding their squares; x² + y² = 119 subtracts the squares, 144 − 25, treating 12 as the hypotenuse of the right-angled triangle rather than as one of the shorter sides.
- (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.
- (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) (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.
- (b) 25 — The tangent at (9, 12) is 9x + 12y = 225. Setting y = 0 (the x-axis): 9x = 225, so x = 25. Choosing 18.75 comes from swapping the coefficients in the tangent equation (using 12x + 9y = 225) before setting y = 0. Choosing 15 is where the circle itself meets the x-axis (from x² = 225), not where the tangent does. Choosing 9 is just the x-coordinate of the original point (9, 12), not the point P.
- (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.
- (b) −3/4 — The tangent to a circle at a point is always perpendicular to the radius drawn to that point. The radius from (0, 0) to (3, 4) has gradient 4/3. The gradient of a line perpendicular to a line with gradient m is the negative reciprocal, −1/m, so the tangent's gradient here is −3/4. Using the radius's own gradient, forgetting that the tangent is perpendicular to it, gives 4/3. Negating the radius's gradient without also taking the reciprocal gives −4/3. Taking the reciprocal of the radius's gradient without negating it gives 3/4.
- (c) 50 — Method: the edge of the track is a circle centred on the origin, so it has equation x² + y² = r²; use the floodlight whose coordinates are given to find r², then substitute x = 0 to find where that circle meets the positive y-axis. Working: 30² + 40² = 900 + 1600 = 2500, so the edge of the track is x² + y² = 2500. Substituting x = 0 gives k² = 2500, so k = √2500 = 50, taking the positive root because k is positive. Answer: k = 50. The distractors: 2500 is r² used as though it were the radius, which would place the second floodlight fifty times too far from the centre; 70 adds the two coordinates, 30 + 40, instead of adding their squares; 40 takes the y-coordinate of the first floodlight to be the radius, which would only be true if that floodlight stood on the y-axis itself.
- (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) (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.
- (a) Outside: OQ² = 221 > r² = 169 — OQ² = 10² + 11² = 100 + 121 = 221. Comparing this with r² = 169: since 221 > 169, OQ > r, so Q lies outside the circle — this is the correct verdict AND the correct working. 'Outside: OQ = 21 (10 + 11) > r = 13' reaches the same Outside verdict, but by invalid working: it adds the coordinates instead of squaring them (10 + 11 = 21, rather than 10² + 11² = 221), so the stated 'OQ' of 21 is not a distance at all — the verdict happens to match, but the method is wrong. 'Inside: OQ ≈ 14.87 < r² = 169' correctly finds the distance OQ = √221 ≈ 14.87, but then compares that DISTANCE with r² = 169 instead of with r = 13 — comparing two different kinds of quantity gives a meaningless, and here wrong, verdict. 'Inside: OQ² = 221 < (2r)² = 676' confuses the radius with the diameter: it compares OQ² with the diameter squared, (2 × 13)² = 676, instead of with r² = 169.
Build your own mix at the worksheet builder.