For what positive integral value of C the points $(2, - 4), (6, 4), (5, 5)$ and $(C, 1)$ are con cyclic

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Solution

Let the equation of the circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
Since it passes through $(2, - 4), (6, 4), (5, 5)$
$4 g - 8 f + c = - 20 . . . . . (1)$
$12 g + 8 f + c = - 52 . . . . (2)$
$10 g + 10 f + c = - 50 . . . . (3)$
Solving (1) and (2) , we get
$8 g + c = - 36$ ....(4)
Solving (1) and (3) , we get
$60 g + 9 c = - 300$
$\Rightarrow 20 g + 3 c = - 100$ ....(5)
Solving (4) and (5), we get
$∴ 4 g = - 8 \Rightarrow g = - 2, c = - 20, f = - 1.$
$∴$ The equation of the circle is
$x^{2} + y^{2} - 4 x - 2 y - 20 = 0$

$(C, 1)$ lies on this circle, then
$\Rightarrow C^{2} - 4 C - 21 = 0$
$(C - 7) (C + 3) = 0$
$C = 7 o r - 3$ .

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