If theta is the angle subtended at P(x_1, y_1 | vedclass

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(B) Let $C$ be the center of the circle and $T_1$ be the point of tangency from $P$. The center $C$ is $(-g, -f)$ and the radius $r = \sqrt{g^2 + f^2

Vedclass · Accepted Answer

$\cot \frac{	heta}{2} = \frac{\sqrt{S_1}}{\sqrt{g^2 + f^2 - c}}$

Vedclass · Answer

(B) Let $C$ be the center of the circle and $T_1$ be the point of tangency from $P$. The center $C$ is $(-g, -f)$ and the radius $r = \sqrt{g^2 + f^2 - c}$.In the right-angled triangle $	riangle PT_1C$,the angle at $P$ is $\frac{	heta}{2}$.Thus,$\cot \frac{	heta}{2} = \frac{PT_1}{CT_1}$.Here,$PT_1$ is the length of the tangent from $P$ to the circle,which is $\sqrt{S_1} = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$.$CT_1$ is the radius $r = \sqrt{g^2 + f^2 - c}$.Therefore,$\cot \frac{	heta}{2} = \frac{\sqrt{S_1}}{\sqrt{g^2 + f^2 - c}}$.

If $\theta$ is the angle subtended at $P(x_1, y_1)$ by the circle $S \equiv x^2 + y^2 + 2gx + 2fy + c = 0$,then

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