O-Level Coordinate geometry in two dimensions: worked solution

Question

For circle C1: x² + y² − 6x + 4y = 12 (centre (3,−2), radius 5), find the tangent at P(7, −5). [3]

The key circle property is that the tangent at a point is PERPENDICULAR to the radius drawn to that point. So find the gradient of the radius first, then turn it into the tangent gradient. (Check P really is on the circle: 7² + (-5)² - 6(7) + 4(-5) = 49 + 25 - 42 - 20 = 12, which matches the right-hand side, so P lies on C1.) The radius joins the centre (3, -2) to P(7, -5). Gradient of a line through (x1, y1) and (x2, y2) is (y2 - y1)/(x2 - x1): gradient of radius = (-5 - (-2)) / (7 - 3) = (-5 + 2) / 4 = -3/4. For two perpendicular lines the gradients multiply to -1, so if the tangent gradient is m: (-3/4) × m = -1 m = -1 ÷ (-3/4) m = 4/3. Now use the point-gradient form y - y1 = m(x - x1) with the point P(7, -5) and m = 4/3: y - (-5) = (4/3)(x - 7) y + 5 = (4/3)(x - 7). Multiply both sides by 3 to clear the fraction: 3(y + 5) = 4(x - 7) 3y + 15 = 4x - 28 3y = 4x - 28 - 15 3y = 4x - 43.