The polars drawn from (-1, 2) to the circles | vedclass

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(D) The equation of the polar of a point $(x_1, y_1)$ with respect to a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $xx_1 + yy_1 + g(x + x_1) +

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Intersect at a point

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(D) The equation of the polar of a point $(x_1, y_1)$ with respect to a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0$. For circle $S_1: x^2 + y^2 + 6y + 7 = 0$,with point $(-1, 2)$,the polar is: $x(-1) + y(2) + 3(y + 2) + 7 = 0$ $-x + 2y + 3y + 6 + 7 = 0$ $-x + 5y + 13 = 0$ or $x - 5y - 13 = 0$. For circle $S_2: x^2 + y^2 + 6x + 1 = 0$,with point $(-1, 2)$,the polar is: $x(-1) + y(2) + 3(x - 1) + 1 = 0$ $-x + 2y + 3x - 3 + 1 = 0$ $2x + 2y - 2 = 0$ or $x + y - 1 = 0$. Comparing the slopes: $m_1 = 1/5$ and $m_2 = -1$. Since $m_1 
eq m_2$ and $m_1 	imes m_2 
eq -1$,the lines are neither parallel nor perpendicular. Therefore,they must intersect at a point.

The polars drawn from $(-1, 2)$ to the circles $S_1 \equiv x^2 + y^2 + 6y + 7 = 0$ and $S_2 \equiv x^2 + y^2 + 6x + 1 = 0$ are:

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