Given two circles $x^{2} + y^{2} + 2 x - 2 y + 1 = 0$ and $x^{2} + y^{2} + 4 x + 4 y + 3 = 0$ . Find the equation of the circle that is orthogonal to both these circle with centre on $x - y = 1$ .

x^{2} + y^{2} + x - y - 3 = 0

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x^{2} - y^{2} - x + y - 3 = 0

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x^{2} + y^{2} + 2 x - y + 3 = 0

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none of these

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Solution

The correct option is A $x^{2} + y^{2} + x - y - 3 = 0$
$x^{2} + y^{2} + 2 x - 2 y + 1 = 0$ ......(1)
$x^{2} + y^{2} + 4 x + 4 y + 3 = 0$ ......(2)

Let the required equation be $x^{2} + y^{2} + 2 f x + 2 g y + c = 0$ .......(3)

Since it is orthogonal to both 1 and 2, we have

$2 g - 2 f = c + 1$

$4 g + 4 f = c + 3$

Also $(- g, - f)$ lies on the $x - y = 1$

$g - f = - 1$

$c = - 3$

$g = - \frac{1}{2}$

$f = \frac{1}{2}$

Therefore, required equation is $x^{2} + y^{2} + x - y - 3 = 0$ .

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