The inverse point of (1,2) with respect to th | vedclass

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(C) The inverse point of $P(x_1, y_1)$ with respect to a circle $S=0$ is the foot of the perpendicular from the center of the circle to the polar of $

Vedclass · Accepted Answer

$(0,1)$

Vedclass · Answer

(C) The inverse point of $P(x_1, y_1)$ with respect to a circle $S=0$ is the foot of the perpendicular from the center of the circle to the polar of $P$ with respect to the circle,or more generally,the point $P'$ such that $P, P'$ are collinear with the center and $CP \cdot CP' = r^2$.Given circle: $x^2+y^2-4x-6y+9=0$. Center $C = (2,3)$,radius $r = \sqrt{2^2+3^2-9} = \sqrt{4+9-9} = 2$.Polar of $P(1,2)$ w.r.t. the circle is $x(1)+y(2)-2(x+1)-3(y+2)+9=0$.Simplifying: $x+2y-2x-2-3y-6+9=0$ $\Rightarrow -x-y+1=0$ $\Rightarrow x+y-1=0$.The inverse point $P'(x', y')$ lies on the line $CP$ and satisfies $CP \cdot CP' = r^2$.Line $CP$ passes through $(2,3)$ and $(1,2)$,so its equation is $y-3 = \frac{2-3}{1-2}(x-2)$ $\Rightarrow y-3 = x-2$ $\Rightarrow y=x+1$.Intersection of $x+y-1=0$ and $y=x+1$: $x+(x+1)-1=0$ $\Rightarrow 2x=0$ $\Rightarrow x=0, y=1$.Thus,the inverse point is $(0,1)$.

The inverse point of $(1,2)$ with respect to the circle $x^2+y^2-4x-6y+9=0$ is

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