Find the equation of the circle the end points of whose diameter are the centres of the circles $x^{2} + y^{2} + 16 x - 14 y - 1 = 0$ & $x^{2} + y^{2} - 4 x + 10 y - 2 = 0$

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Solution

Centre of first circle
$x^{2} + y^{2} + 16 x - 14 y - 1 = 0. i s (- g, - f) \Rightarrow (- 8, 7) \to P F o r I I C i r c l e : C e n t r e \to (2, - 5) \to Q E q u a t i o n o f c i r c l e w i t h P & Q a s d i a m e t r i c e n d p o i n t s : C e n t r e \to (\frac{- 8 + 2}{2}, \frac{7 - 5}{2}) \Rightarrow (- 3, 1) . r a d i u s = \frac{1}{2} \sqrt{(2 + 8)^{2} + (- 5 - 7)^{2}} = \frac{1}{2} \sqrt{100 + 144} = \frac{\sqrt{244}}{2} . \Rightarrow (x + 3)^{2} + (y - 1)^{2} = \frac{244}{4} \Rightarrow (x + 3)^{2} + (y - 1)^{2} = 61 A n s .$

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