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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- (d) y = −2x + 10 — Method: a tangent is perpendicular to the radius drawn to the point where it touches, so work out the gradient of that radius, take its negative reciprocal for the tangent, then substitute into y − y₁ = m(x − x₁). Working: the radius joins (0, 0) to (4, 2), so its gradient is 2 ÷ 4 = 1/2; turning 1/2 upside down gives 2 and changing the sign gives −2. Substituting into y − 2 = −2(x − 4) gives y − 2 = −2x + 8, so y = −2x + 10. Answer: y = −2x + 10. The distractors: y = −0.5x + 4 changes the sign of the radius gradient but never turns it upside down, using −1/2 where −2 belongs; y = 2x − 6 turns the gradient upside down but leaves it positive, using 2 where −2 belongs; y = −2x − 10 has the correct gradient but substitutes the point with both signs reversed, writing y + 2 = −2(x + 4) instead of y − 2 = −2(x − 4).
- (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.
- (c) y = (5/12)x − 169/12 — The gradient of the radius to (5, −12) is (−12 − 0) ÷ (5 − 0) = −12/5. A tangent is perpendicular to the radius at that point, so its gradient is the negative reciprocal, 5/12. Using y − y₁ = m(x − x₁) with (5, −12): y + 12 = (5/12)(x − 5), which gives y = (5/12)x − 169/12. y = −(12/5)x comes from using the radius's own gradient, −12/5, instead of turning it into the perpendicular gradient, and building the line through the origin (as the radius itself does). y = −(5/12)x − 119/12 comes from taking the reciprocal of −12/5 correctly as a size but keeping the wrong sign, using −5/12 instead of 5/12. y = (5/12)x − 25/12 comes from using the correct gradient 5/12 but building the line through (5, 0) instead of (5, −12) — dropping the point's y-coordinate.
- (a) −3/4 — Method: the tangent at a point on a circle is perpendicular to the radius drawn to that point, so find the gradient of the radius and then take its negative reciprocal. Working: the radius joins (0, 0) to (6, 8), so its gradient is 8 ÷ 6, which cancels to 4/3. Turning 4/3 upside down gives 3/4, and changing the sign gives −3/4. Answer: the gradient of the tangent is −3/4. The distractors: 4/3 is the gradient of the radius itself, quoted without taking the perpendicular at all; −4/3 changes the sign but leaves the fraction the same way up, so the two gradients do not multiply to give −1; 3/4 turns the fraction upside down but keeps it positive, which is the other half of the same rule left undone.
- (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.
- (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.
- (a) (6, 8) — Method: a point lies on the circle x² + y² = 100 exactly when the squares of its two coordinates add to 100, so square both coordinates of each point and add them. Working: for (6, 8), 6² + 8² = 36 + 64 = 100, which matches the right-hand side of the equation. Answer: (6, 8) lies on the circle. The distractors: (3, 4) is the 3, 4, 5 right-angled triangle recalled but never scaled up to a radius of 10, and 3² + 4² = 25, so it lies on the far smaller circle x² + y² = 25; (5, 5) has coordinates adding to 10, which compares the sum of the coordinates with the radius instead of the sum of their squares with r², and 5² + 5² = 50; (10, 10) takes each coordinate separately to equal the radius, and 10² + 10² = 200, which is twice too big.
- (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) 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.
- (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.
- (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.
- (d) P lies outside the circle — The squared distance from the origin to P is 8² + 9² = 64 + 81 = 145, which is greater than 100, so P lies outside the circle. Choosing 'P lies inside the circle' comes from adding the coordinates directly (8 + 9 = 17) instead of squaring and adding them, and comparing 17 with 100. Choosing 'P lies on the circle' comes from squaring 9 incorrectly as 36 (as if it were 6²) instead of 81, giving 64 + 36 = 100, equal to the right-hand side. Choosing 'Cannot be determined' wrongly assumes the equation of the line itself is needed, when the radius and the distance from the origin to P are already enough.
- (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.
- (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) Yes, because 3² + 4² = 25. — A point lies on the circle x² + y² = 25 exactly when substituting its coordinates makes the equation true. Squaring each coordinate separately and adding: 3² + 4² = 9 + 16 = 25, which matches the right-hand side, so (3, 4) does lie on the circle. Adding the coordinates without squaring them, 3 + 4 = 7, and then reasoning that 7 is less than 25 happens to reach the same verdict, but it is not testing the equation of the circle at all — the circle equation depends on x² + y², not x + y. Squaring the sum instead of summing the squares, (3 + 4)² = 49, not 25, wrongly rules the point out. Doubling each coordinate instead of squaring it, so that 4² is taken as 8, gives 9 + 8 = 17, not 25, which also wrongly rules the point out.
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