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
Equation of a circle and its tangent worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- 1.A coastguard radar at the origin covers a circular region modelled by x² + y² = 400, where each unit represents 1 kilometre. A boat travels along the straight line that touches the boundary of the region at the point (12, 16). Work out the equation of the line the boat travels along.
- 2.The point (9, 12) lies on the circle x² + y² = 225, which has centre (0, 0). The tangent to the circle at (9, 12) crosses the x-axis at the point P. Work out the x-coordinate of P.
- 3.A circle has equation x² + y² = 49. Work out the coordinates of the point(s) on the circle where the tangent is horizontal.
- 4.A circle has centre (0, 0) and equation x² + y² = 100. Work out which one of these points lies on the circle.
- 5.A circle has centre (0, 0) and passes through the point (5, 12). Work out the equation of the circle.
- 6.The point (20, 21) lies on the circle x² + y² = 841, which has centre O(0, 0). The tangent to the circle at (20, 21) crosses the x-axis at P and the y-axis at Q. Work out the area of triangle OPQ, correct to 1 decimal place.
- 7.A circle has equation x² + y² = 25. Does the point (3, 4) lie on this circle?
- 8.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.
- 9.A circular running track is modelled on a grid whose centre is the origin, where each unit represents 1 metre. A floodlight at the point (30, 40) stands on the edge of the track. A second floodlight stands on the edge of the track at the point (0, k), where k is positive. Work out the value of k.
- 10.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.
- 11.The point A(−6, 8) lies on the circle x² + y² = 100, whose centre is the origin O. The tangent to the circle at A crosses the y-axis at the point B. Work out the length of OB.
- 12.The point (3, 6) lies on the circle x² + y² = 45. The tangent to the circle at (3, 6) crosses the x-axis at the point P. Work out the coordinates of P.
- 13.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.
- 14.A circle has centre (0, 0) and equation x² + y² = 50. Work out the radius of the circle, giving your answer as a surd in its simplest form.
- 15.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.
Answer key
- (b) y = −3x/4 + 25 — Method: a straight line that touches a circle at one point is a tangent there, so it is perpendicular to the radius drawn to that point; find the gradient of the radius, take its negative reciprocal, then substitute the point of contact into y − y₁ = m(x − x₁). Working: the radius from (0, 0) to (12, 16) has gradient 16 ÷ 12, which cancels to 4/3, so the tangent has gradient −3/4. Substituting gives y − 16 = −3/4(x − 12), so y − 16 = −3x/4 + 9 and y = −3x/4 + 25. Answer: y = −3x/4 + 25. The distractors: y = 3x/4 + 7 turns the gradient of the radius upside down but leaves it positive, so the perpendicular step is only half done; y = −4x/3 + 32 changes the sign of the radius gradient without turning it upside down, which is the other half left undone; y = −3x/4 − 25 uses the correct gradient but substitutes the point of contact with both signs reversed, writing y + 16 = −3/4(x + 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.
- (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) (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.
- (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.
- (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.
- (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.
- (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.
- (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) 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.
- (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.
- (d) 5√2 — For a circle x² + y² = r², the 50 on the right-hand side is r², not r, so the radius is √50. Writing 50 as 25 × 2, the largest square factor times what remains, gives √50 = √25 × √2 = 5√2. Forgetting to square-root 50 at all and giving the value of r² instead gives 50. Halving 50 instead of taking its square root gives 25. Using 25 as the number left outside the square root sign, instead of as the number under it, gives the wrongly simplified 25√2.
- (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.
Build your own mix at the worksheet builder.