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9709 P11 - Jun 2020 - Q10
639
The coordinates of the points A and B are
(-1, -2) and (7, 4) respectively.
(a) Find the equation of the circle, C, for which AB is a diameter.
(b) Find the equation of the tangent, T, to circle C at the point B.
(c) Find the equation of the circle which is the reflection of circle C in the line T.
Solution
(a) The midpoint of AB is the center of the circle C.
Calculate the midpoint:
\(\left(\frac{-1+7}{2}, \frac{-2+4}{2}\right) = (3, 1)\).
The radius is half the distance of AB.
Calculate the distance:
\(\sqrt{(7 - (-1))^2 + (4 - (-2))^2} = \sqrt{64 + 36} = 10\).
Thus, the radius is 5.
The equation of the circle is
\((x-3)^2 + (y-1)^2 = 5^2 = 25\).
(b) The gradient of AB is
\(\frac{4 - (-2)}{7 - (-1)} = \frac{6}{8} = \frac{3}{4}\).
The gradient of the tangent T is the negative reciprocal:
\(-\frac{4}{3}\).
The equation of the tangent at B is
\(y - 4 = -\frac{4}{3}(x - 7)\).
Simplifying gives
\(3y + 4x = 40\).
(c) The reflection of the circle C in the line T has its center at
\((11, 7)\) (since B is the midpoint of the line joining the centers).
The radius remains the same, so the new equation is
\((x-11)^2 + (y-7)^2 = 25\).
