The poles of the tangents to the circle x^2+y | vedclass

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Question

(A) Let the tangent to the circle $x^2+y^2=4$ be $x \cos 	heta + y \sin 	heta = 2$. This line can be written as $x \cos 	heta + y \sin 	heta - 2 =

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$y^2+8x=0$

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(A) Let the tangent to the circle $x^2+y^2=4$ be $x \cos 	heta + y \sin 	heta = 2$. This line can be written as $x \cos 	heta + y \sin 	heta - 2 = 0$. Let $(x_1, y_1)$ be the pole of this tangent with respect to the circle $(x+2)^2+y^2=8$,which expands to $x^2+4x+4+y^2=8$ or $x^2+y^2+4x-4=0$. The equation of the polar of $(x_1, y_1)$ with respect to $x^2+y^2+4x-4=0$ is given by $x x_1 + y y_1 + 2(x+x_1) - 4 = 0$. Rearranging this,we get $(x_1+2)x + y_1 y + (2x_1-4) = 0$. Since this represents the same line as the tangent,we compare the coefficients: $\frac{\cos 	heta}{x_1+2} = \frac{\sin 	heta}{y_1} = \frac{-2}{2x_1-4} = \frac{-1}{x_1-2}$. Thus,$\cos 	heta = -\frac{x_1+2}{x_1-2}$ and $\sin 	heta = -\frac{y_1}{x_1-2}$. Using the identity $\cos^2 	heta + \sin^2 	heta = 1$,we have $\left(-\frac{x_1+2}{x_1-2}ight)^2 + \left(-\frac{y_1}{x_1-2}ight)^2 = 1$. This simplifies to $(x_1+2)^2 + y_1^2 = (x_1-2)^2$. $x_1^2 + 4x_1 + 4 + y_1^2 = x_1^2 - 4x_1 + 4$. $y_1^2 + 8x_1 = 0$. Replacing $(x_1, y_1)$ with $(x, y)$,the locus is $y^2+8x=0$.

The poles of the tangents to the circle $x^2+y^2=4$ with respect to the circle $(x+2)^2+y^2=8$ lie on

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