If the polar of a point on the circle x^2+y^2 | vedclass

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(B) Let the point on the circle $x^2+y^2=p^2$ be $(x_1, y_1)$.Then $x_1^2+y_1^2=p^2$.The equation of the polar of $(x_1, y_1)$ with respect to the cir

Vedclass · Accepted Answer

$GP$

Vedclass · Answer

(B) Let the point on the circle $x^2+y^2=p^2$ be $(x_1, y_1)$.Then $x_1^2+y_1^2=p^2$.The equation of the polar of $(x_1, y_1)$ with respect to the circle $x^2+y^2=q^2$ is $x x_1+y y_1=q^2$.This line touches the circle $x^2+y^2=r^2$.The perpendicular distance from the center $(0, 0)$ to the line $x x_1+y y_1-q^2=0$ must be equal to the radius $r$.So,$\frac{|0(x_1)+0(y_1)-q^2|}{\sqrt{x_1^2+y_1^2}} = r$.$|q^2| = r \sqrt{x_1^2+y_1^2}$.Since $x_1^2+y_1^2=p^2$,we have $q^2 = r \sqrt{p^2} = rp$.Thus,$q^2 = pr$,which implies that $p, q, r$ are in $GP$.

If the polar of a point on the circle $x^2+y^2=p^2$ with respect to the circle $x^2+y^2=q^2$ touches the circle $x^2+y^2=r^2$,then $p, q, r$ are in

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