The common chord of the circles $x^{2} + y^{2} + 4 x + 1 = 0$ and $x^{2} + y^{2} + 6 x + 2 y + 3 = 0$ is

x + y + 1 = 0

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5 x + y + 2 = 0

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2 x + 2 y + 5 = 0

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3 x + y + 3 = 0

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Solution

The correct option is A $x + y + 1 = 0$
$\begin{matrix} x^{2} + y^{2} + 6 x + 2 y + 3 = 0 x^{2} + y^{2} + 4 x + 1 = 0 S u b t r a c t i n g t h e a b o v e e q u a t i o n s w e g e t c o m m o n c h o r d : 2 x + 2 y + 2 = 0 x + y + 1 = 0 H e n c e, t h e o p t i o n A i s t h e c o r r e c t a n s w e r . \end{matrix}$

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Similar questions

The common chord of the circle $x^{2} + y^{2} + 4 x + 1 = 0$ and $x^{2} + y^{2} + 6 x + 2 y + 3 = 0$ is

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 + 1 = 0; x^{2} + y^{2} + 4 x + 3 y + 2 = 0

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

120^{\circ}

2 x - y - 2 = 0

(3, - 2)

(- 2, 0)

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