Question

If the distance of the chord of contact of a circle from any point on its director circle is $\alpha$, then the radius of the circle is

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

Answer 1

The correct option is C $\sqrt{2}\alpha$

If $S_1:(x–a{)}^2+(y–b{)}^2=2r^2$

$CP=\sqrt{2}r,CA=r$

$\angle CAP={90}^o$

$⟹C{P}^2=C{A}^2+A{P}^2$

$⟹AP=r$

$\tan \theta =\frac{CA}{AP}=\frac{r}{r}=1$

$\theta ={45}^{\circ }$

In $△APQ$

$\angle AQP={90}^o$

$\cos \theta =\frac{PQ}{AP}$

$AP=r=\frac{\alpha }{\cos {45}^o}$

$⟹r=\sqrt{2}\alpha$