Question

If $\theta$ is the angle subtended by the circle $S=x^2+y^2+2gx+2fy=c$ at the point $P(x_1,y_1)$ and $S_1=x_1^2+y_1^2+2g{x}_1+2f{y}_1+c,$ then

• A. $\cot \theta =\frac{2\sqrt{S_1}}{\sqrt{g^2+f^2-c}}$
• B. $\cot \frac{\theta }{2}=\frac{\sqrt{S_1}}{\sqrt{g^2+f^2-c}}$
• C. $\theta =2{\tan }^{-1}\sqrt{\frac{g^2+f^2-c}{S_1}}$
• D. $\cos \theta =\frac{S_1-c-g^2-f^2}{S_1-c+g^2+f^2}$

Answer 1

The correct option is B $\cot \frac{\theta }{2}=\frac{\sqrt{S_1}}{\sqrt{g^2+f^2-c}}$

WKT OA =$r$ , O =$(-g,-f)$ and $\angle OAP={90}^o$

$⟹A{P}^2=O{P}^2–r^2$

$=\left(\right(x_1+g{)}^2+(y_1+f{)}^2)–(\sqrt{g^2+f^2-c}{)}^2$

$A{P}^2=x_1^2+y_1^2+2g{x}_1+2f{y}_1+c=S_1$

$AP=\sqrt{S_1}$

In $△AOP$

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