Prove that the circle through the origin and cutting the circles
$x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} = 0$ and $x^{2} + y^{2} + 2 g_{2} x + 2 f_{2} y + c_{2} = 0$
orthogonally is $∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} + y^{2} & - x & - y c_{1} & g_{1} & f_{1} c_{2} & g_{2} & f_{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$

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Solution

Let the circle through origin be
$x^{2} + y^{2} + 2 g x + 2 f y = 0$
Condition of orthogonality gives
$2 g g_{1} + 2 f f_{1} = 0 + c_{1} . . . . . . (1)$
$2 g g_{2} + 2 f f_{2} = 0 + c_{2} . . . . (2)$
The given circle can be written as
$2 g x + 2 f y = - (x^{2} + y^{2}) . . . . . (3)$
Eliminating the unknown quantities $g$ and $f$ from $(1), (2)$ and $(3)$ , we get the result in determinant form.

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Two circles $x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} = 0 a n d x^{2} + y^{2} + 2 g_{2} x + 2 f_{2} y + c^{2} = 0$ are said to be orthogonal. Then $2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$

Q. Find the equation of a circle passing through the origin and cutting the circles

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x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y = 0

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x^{2} + y^{2} + 2 g_{2} x + 2 f_{2} y = 0

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f_{1} g_{2} = f_{2} g_{1}

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Prove that the circle through the origin and cutting the circlesx2+y2+2g1x+2f1y+c1=0 and x2+y2+2g2x+2f2y+c2=0orthogonally is ∣∣ ∣ ∣∣x2+y2−x−yc1g1f1c2g2f2∣∣ ∣ ∣∣=0

Let the circle through origin bex2+y2+2gx+2fy=0Condition of orthogonality gives2gg1+2ff1=0+c1......(1)2gg2+2ff2=0+c2....(2)The given circle can be written as2gx+2fy=−(x2+y2).....(3)Eliminating the unknown quantities g and f from (1),(2) and (3), we get the result in determinant form.