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

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(C) The inverse point of a point $P(x_1, y_1)$ with respect to a circle $S = 0$ is the point $P'$ such that $P'$ lies on the line joining the center o

Vedclass · Accepted Answer

$(0, 1)$

Vedclass · Answer

(C) The inverse point of a point $P(x_1, y_1)$ with respect to a circle $S = 0$ is the point $P'$ such that $P'$ lies on the line joining the center of the circle to $P$,and the product of their distances from the center is equal to the square of the radius.Alternatively,the inverse point is the foot of the perpendicular from the center of the circle to the polar of $P$ if $P$ is outside,or more generally,the point $P'$ such that the polar of $P'$ is the line passing through $P$ perpendicular to the line joining the center to $P$.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 $P(1, 2)$ with respect to the circle is $x(1) + y(2) - 2(x + 1) - 3(y + 2) + 9 = 0$.$x + 2y - 2x - 2 - 3y - 6 + 9 = 0$ $\Rightarrow -x - y + 1 = 0$ $\Rightarrow x + y - 1 = 0$.The inverse point $P'(x', y')$ lies on the line passing through $C(2, 3)$ and $P(1, 2)$. The slope of $CP$ is $\frac{3-2}{2-1} = 1$. The equation of line $CP$ is $y - 2 = 1(x - 1) \Rightarrow y = x + 1$.Intersection of $x + y - 1 = 0$ and $y = x + 1$: $x + (x + 1) - 1 = 0$ $\Rightarrow 2x = 0$ $\Rightarrow x = 0$. Then $y = 1$.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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