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June 2023 p11 q12
644
The diagram shows a circle P with centre (0, 2) and radius 10 and the tangent to the circle at the point A with coordinates (6, 10). It also shows a second circle Q with centre at the point where this tangent meets the y-axis and with radius \(\frac{5}{2} \sqrt{5}\).
(a) Write down the equation of circle P.
(b) Find the equation of the tangent to the circle P at A.
(c) Find the equation of circle Q and hence verify that the y-coordinates of both of the points of intersection of the two circles are 11.
(d) Find the coordinates of the points of intersection of the tangent and circle Q, giving the answers in surd form.
Solution
(a) The equation of a circle with center (h, k) and radius r is \((x-h)^2 + (y-k)^2 = r^2\). For circle P, the center is (0, 2) and the radius is 10, so the equation is \(x^2 + (y-2)^2 = 100\).
(b) The gradient of the radius at point A is \(\frac{10-2}{6-0} = \frac{4}{3}\). The gradient of the tangent is the negative reciprocal, \(-\frac{3}{4}\). Using point-slope form, the equation of the tangent at (6, 10) is \(y - 10 = -\frac{3}{4}(x - 6)\), which simplifies to \(y = \frac{3}{4}x + \frac{29}{2}\).
(c) The center of circle Q is where the tangent meets the y-axis, which is \(\left(0, \frac{29}{2}\right)\). The radius is \(\frac{5}{2} \sqrt{5}\), so the equation is \(x^2 + \left(y - \frac{29}{2}\right)^2 = \left(\frac{5}{2} \sqrt{5}\right)^2 = 31.25\). To verify the y-coordinates of intersection, solve \(x^2 + (11-2)^2 = 100\) and \(x^2 + (11-11)^2 = 31.25\), confirming y-coordinates are 11.
(d) Substitute the tangent equation into circle Q's equation: \(x^2 + \left(\frac{3}{4}x + \frac{29}{2} - \frac{29}{2}\right)^2 = 31.25\). Solving gives \(x = \pm \sqrt{20}\). For each x, find y: \(y = \frac{3}{4}x + \frac{29}{2}\), resulting in \(y = \frac{58 \pm 3\sqrt{20}}{4}\).
