Question

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

• A. $\cot \alpha =\frac{\sqrt{S_1}}{\sqrt{g^2+f^2-c}}$
• B. $\cot \alpha /2=\frac{\sqrt{S_1}}{\sqrt{g^2+f^2-c}}$
• C. $\tan \alpha =\frac{2\sqrt{g^2+f^2-c}}{\sqrt{S_1}}$
• D. $\alpha =2{\tan }^{-1}\left(\frac{\sqrt{g^2+f^2-c}}{\sqrt{S_1}}\right)$

Answer 1

Answer: (B), (D)

$\cot \alpha /2=\frac{\sqrt{S_1}}{\sqrt{g^2+f^2-c}}$
$\alpha =2{\tan }^{-1}\left(\frac{\sqrt{g^2+f^2-c}}{\sqrt{S_1}}\right)$

where $S_1=x_1^2+y_1^2+2g{x}_1+2f{y}_1+c$ Ans. (b) and (d)
Solution Let $PA$ and $PB$ he the tangents from $P(x_1,y_1)$ to the given circle with centre $C(-g-f),$ such that $\angle APB=\theta$. Then $\angle APC=\theta /2(Fig.16.31).$ Therefore, from $\Delta .PAC,$ we get
$\cot =\frac{\theta }{2}=\frac{PA}{AC}=\frac{\sqrt{S_1}}{\sqrt{g^2+f^2-c}}$
$\Rightarrow \tan =\frac{\theta }{2}=\frac{\sqrt{g^2+f^2-c}}{\sqrt{S_1}}$
$\Rightarrow \theta =2{\tan }^{-1}\left(\frac{\sqrt{g^2+f^2-c}}{\sqrt{S_1}}\right)$