Find the equation of circle passing through the points where the circles
$x^{2} + y^{2} + 6 x - 8 y - 11 = 0$ and
$x^{2} + y^{2} - 8 x + 14 y + 56 = 0$ subtend equal angles and cut the first of these circles orthogonally.

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Solution

The given circles are
$C_{1} (- 3, 4), r_{1} = 6$
$C_{2} (4, - 7), r_{2} = 3$
The circles subtend equal angles at the points from where common tangents are drawn. These points divide the line of centres in the ratio of the radii internally and externally. These are easily found to be $A (11, - 18), B (\frac{5}{3}, - \frac{10}{3})$ .
If the required circle be
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
Since it passes through the points $A$ and $B$ , we have
$22 g - 36 f + c = - 445 . . . . . . (1)$
$\frac{10}{3} g - \frac{20}{3} f + c = - \frac{125}{9} . . . . . (2)$
Also condition of orthogonality with the first gives
$6 g - 8 f - c = - 11$
Adding $(1)$ and $(3)$ and then $(2)$ and $(3)$ thus eliminating $c$ , we get
$7 g - 11 f = - 144$ and $7 g - 11 f = - \frac{56}{3}$
These are inconsistent. Hence no such circle exists.

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