If the length of the tangent from (h, k) to t | vedclass

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(C) The length of the tangent from a point $(h, k)$ to a circle $x^2+y^2+2gx+2fy+c=0$ is given by $\sqrt{h^2+k^2+2gh+2fk+c}$.For the circle $x^2+y^2-1

Vedclass · Accepted Answer

$3h^2+3k^2+8h+8k+16=0$

Vedclass · Answer

(C) The length of the tangent from a point $(h, k)$ to a circle $x^2+y^2+2gx+2fy+c=0$ is given by $\sqrt{h^2+k^2+2gh+2fk+c}$.For the circle $x^2+y^2-16=0$,the length of the tangent is $L_1 = \sqrt{h^2+k^2-16}$.For the circle $x^2+y^2+2x+2y=0$,the length of the tangent is $L_2 = \sqrt{h^2+k^2+2h+2k}$.According to the problem,$L_1 = 2L_2$.Squaring both sides,we get $L_1^2 = 4L_2^2$.$h^2+k^2-16 = 4(h^2+k^2+2h+2k)$.$h^2+k^2-16 = 4h^2+4k^2+8h+8k$.Rearranging the terms,we get $3h^2+3k^2+8h+8k+16=0$.

If the length of the tangent from $(h, k)$ to the circle $x^2+y^2=16$ is twice the length of the tangent from the same point to the circle $x^2+y^2+2x+2y=0$,then

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