Question

If $\theta$ is the angle between the tangents from $(0,0)$ to the circle $x^2+y^2-14x+2y+25=0$ then $\sin \left(\frac{\theta }{2}\right)+\cos \left(\frac{\theta }{2}\right)=$

• A. $2\sqrt{2}$
• B. $\frac{1}{2\sqrt{2}}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\sqrt{2}$

Answer 1

The correct option is D $\sqrt{2}$

$S_1:x^2+y^2-14x+2y+25=0$
$C\equiv centre\equiv$ (7, -1) radius=r=5
and L=length of tangent from (0,0) to S
$\therefore L=\sqrt{S_1}=\sqrt{0+0-0+0+25}=5$
$tan\left(\frac{\theta }{2}\right)=\frac{r}{L}=\frac{5}{5}=1$
$\therefore \frac{\theta }{2}={45}^{\circ }$
$\theta ={90}^{\circ }$
$\therefore sin\left(\frac{\theta }{2}\right)+cos\left(\frac{\theta }{2}\right)=sin{45}^{\circ }+cos{45}^{\circ }$
$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}$