The locus of the poles of the tangents to the | vedclass

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(A) Let the pole be $P(x_1, y_1)$. The polar of $P$ with respect to the circle $x^2+y^2=4$ is $xx_1+yy_1=4$.Since this polar is a tangent to the circl

Vedclass · Accepted Answer

$3x^2+3y^2+2xy+8x-8y-16=0$

Vedclass · Answer

(A) Let the pole be $P(x_1, y_1)$. The polar of $P$ with respect to the circle $x^2+y^2=4$ is $xx_1+yy_1=4$.Since this polar is a tangent to the circle $x^2+y^2-2x+2y-2=0$,the perpendicular distance from the center $(1, -1)$ of this circle to the line $xx_1+yy_1-4=0$ must be equal to its radius.The center is $(1, -1)$ and the radius $r = \sqrt{1^2+(-1)^2-(-2)} = \sqrt{4} = 2$.The perpendicular distance is $d = \frac{|1(x_1) + (-1)(y_1) - 4|}{\sqrt{x_1^2+y_1^2}} = 2$.Squaring both sides:$\frac{(x_1-y_1-4)^2}{x_1^2+y_1^2} = 4$$(x_1-y_1-4)^2 = 4(x_1^2+y_1^2)$$x_1^2+y_1^2+16-2x_1y_1-8x_1+8y_1 = 4x_1^2+4y_1^2$$3x_1^2+3y_1^2+2x_1y_1+8x_1-8y_1-16=0$.Replacing $(x_1, y_1)$ with $(x, y)$,the locus is $3x^2+3y^2+2xy+8x-8y-16=0$.

The locus of the poles of the tangents to the circle $x^2+y^2-2x+2y-2=0$ with respect to the circle $x^2+y^2=4$ is:

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