The equation of the normal to the circle $x^{2} + y^{2} + 6 x + 4 y - 3 = 0$ at $(1, - 2)$ is

y + 1 = 0

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y + 2 = 0

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y + 3 = 0

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y - 2 = 0

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Solution

The correct option is C $y + 2 = 0$
Every normal to the circle pass through centre of circle therefore normal to circle . $x^{2} + y^{2} + 6 x + 4 y - 3 = 0 a t (1, - 2)$ also pass through $(- 3, - 2)$
$∴ e q^{n}$ of normal is
$y + 2 = \frac{0}{4} (x - 1)$
$\Rightarrow y + 2 = 0$

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Find the equation of the circle which cuts the following circles orthogonally.
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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

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

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The equation of the normal to the circle x2+y2+6x+4y−3=0 at (1,−2) is

The correct option is C y+2=0Every normal to the circle pass through centre of circle therefore normal to circle . x2+y2+6x+4y−3=0at(1,−2) also pass through (−3,−2)∴eqn of normal isy+2=04(x−1)⇒y+2=0

The equation of the normal to the circle $x^{2} + y^{2} + 6 x + 4 y - 3 = 0$ at $(1, - 2)$ is

The correct option is C $y + 2 = 0$
Every normal to the circle pass through centre of circle therefore normal to circle . $x^{2} + y^{2} + 6 x + 4 y - 3 = 0 a t (1, - 2)$ also pass through $(- 3, - 2)$
$∴ e q^{n}$ of normal is
$y + 2 = \frac{0}{4} (x - 1)$
$\Rightarrow y + 2 = 0$