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9709 P12 - Nov 2023 - Q11
631
The coordinates of points A, B and C are (6, 4), (p, 7) and (14, 18) respectively, where p is a constant. The line AB is perpendicular to the line BC.
(a) Given that p < 10, find the value of p.
A circle passes through the points A, B and C.
(b) Find the equation of the circle.
(c) Find the equation of the tangent to the circle at C, giving the answer in the form dx + ey + f = 0, where d, e and f are integers.
Solution
(a) To find p, use the condition that line AB is perpendicular to line BC. The gradient of AB is \(\frac{7 - 4}{p - 6}\) and the gradient of BC is \(\frac{18 - 7}{14 - p}\). Since they are perpendicular, their product is -1:
Solving this equation gives \(p^2 - 20p + 51 = 0\). Solving the quadratic equation, we find \(p = 3\) or \(p = 17\). Given \(p < 10\), the value of \(p\) is 3.
(b) The center of the circle is the midpoint of AC, which is \((10, 11)\). The radius is the distance from the center to any of the points, say A: \(\sqrt{(10 - 6)^2 + (11 - 4)^2} = \sqrt{65}\). Thus, the equation of the circle is \((x - 10)^2 + (y - 11)^2 = 65\).
(c) The gradient of the radius at C is \(\frac{11 - 18}{10 - 14} = \frac{7}{4}\). The tangent at C is perpendicular to this radius, so its gradient is \(-\frac{4}{7}\). Using point-slope form at C (14, 18), the equation is:
