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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Equation of a circle and its tangent worksheet — GCSE Higher
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- 1.A circle has centre (0, 0) and equation x² + y² = 8. Work out the radius of the circle, giving your answer as a surd in its simplest form.
- 2.A circle has centre (0, 0) and passes through the point (5, 12). Work out the equation of the circle.
- 3.A circle has centre (0, 0) and radius 9. Work out the equation of the circle.
- 4.A circle has centre (0, 0) and passes through the point (12, 35). Work out the equation of the circle.
- 5.A circle has centre (0, 0) and equation x² + y² = 100. A straight line is drawn from the origin through the point P(8, 9). Work out whether P lies inside, on, or outside the circle.
- 6.The point (−4, 3) lies on the circle x² + y² = 25, which has centre (0, 0). Work out the equation of the tangent to the circle at (−4, 3), giving your answer in the form y = mx + c.
- 7.A circle has centre (0, 0) and equation x² + y² = 36. Work out the coordinates of the two points where the circle crosses the y-axis.
- 8.The point (9, 40) lies on the circle x² + y² = 1681, which has centre (0, 0). Work out the equation of the tangent to the circle at (9, 40), giving your answer in the form ax + by = c.
- 9.Points A(8, 15) and B(−8, 15) both lie on the circle x² + y² = 289, which has centre (0, 0). Work out the length of the chord AB.
- 10.The point (6, 8) lies on the circle x² + y² = 100. Work out the gradient of the tangent to the circle at (6, 8).
- 11.A circle has centre (0, 0) and equation x² + y² = 25. Work out the x-coordinates of the two points where the circle crosses the line y = 3.
- 12.The point (3, 4) lies on the circle x² + y² = 25, which has centre (0, 0). Work out the gradient of the tangent to the circle at (3, 4).
- 13.A circle has centre (0, 0) and equation x² + y² = 100. Work out which one of these points lies on the circle.
- 14.The point (5, −12) lies on the circle x² + y² = 169, which has centre (0, 0). Work out the equation of the tangent to the circle at (5, −12), giving your answer in the form y = mx + c.
- 15.A circle has equation x² + y² = 49. Work out the coordinates of the point(s) on the circle where the tangent is horizontal.
Answer key
- (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 = 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 = 81 — The equation of a circle with centre the origin and radius r is x² + y² = r². Here r = 9, so r² = 9 × 9 = 81, giving x² + y² = 81. Choosing x² + y² = 9 uses the radius itself instead of squaring it. Choosing x² + y² = 18 doubles the radius (9 × 2 = 18) instead of squaring it. Choosing x² − y² = 81 keeps the correct 81 but writes a minus instead of a plus, which is not the equation of a circle.
- (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.
- (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) 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) (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) 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.
- (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.
- (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.
- (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.
- (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) (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.
- (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) (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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