The polars of (-1, 2) with respect to the two | vedclass

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(D) The polar of the circle $x^2+y^2+2gx+2fy+c=0$ with respect to the pole $(x_1, y_1)$ is given by $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$. For $S_1 \equiv

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Intersecting at a non-zero point

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(D) The polar of the circle $x^2+y^2+2gx+2fy+c=0$ with respect to the pole $(x_1, y_1)$ is given by $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$. For $S_1 \equiv x^2+y^2+6y+7=0$ with pole $(-1, 2)$,the polar is: $-x+2y+3(y+2)+7=0 \Rightarrow -x+5y+13=0$ (Equation $1$). For $S_2 \equiv x^2+y^2+6x+1=0$ with pole $(-1, 2)$,the polar is: $-x+2y+3(x-1)+1=0$ $\Rightarrow 2x+2y-2=0$ $\Rightarrow x+y-1=0$ (Equation $2$). To check the intersection,we solve the system: $-x+5y+13=0$ $x+y-1=0$ Adding these equations gives $6y+12=0$,so $y=-2$. Substituting $y=-2$ into $x+y-1=0$ gives $x-2-1=0$,so $x=3$. The polars intersect at the point $(3, -2)$,which is a non-zero point.

The polars of $(-1, 2)$ with respect to the two 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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