The locus of the centre of circle $S = 0$ which cuts orthogonally to the circles $x^{2} + y^{2} + 2 x + 3 y + 1 = 0$ and $x^{2} + y^{2} + 3 x + 2 y + 2 = 0$ is

x - y + 1 = 0

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2 x + 2 y + 5 = 0

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x^{2} + y^{2} + x + y - 1 = 0

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

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Solution

The correct option is B $x - y + 1 = 0$
Given Equations of the circle are
$S_{1} : x^{2} + y^{2} + 2 x + 3 y + 1 = 0$
$S_{2} : x^{2} + y^{2} + 3 x + 2 y + 2 = 0$
Now ,Equation of the common chord is
$S_{1} - S_{2} = 0$
$(x^{2} + y^{2} + 2 x + 3 y + 1) - (x^{2} + y^{2} + 3 x + 2 y + 2) = 0$
$- x + y - 1 = 0$
$x - y + 1 = 0$
Hence , Option A

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Similar questions

x^{2} + y^{2} + 2 x + 3 y + 1 = 0, x^{2} + y^{2} + x + y + 2 = 0, x^{2} + y^{2} + 3 x + 2 y + 1 = 0

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

120^{0}

x^{2} + y^{2} + 6 x - 1 = 0

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

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

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

120^{\circ}

S = 0

x^{2} + y^{2} - 2 x + 4 y - 1 = 0

x^{2} + y^{2} + 4 x - 2 y - 5 = 0

x^{2} + y^{2} = 4

S = 0

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