Question

From an arbitrary point 'P' on the cirlce $x^2+y^2=9$. tangents are drawn to the cirlce $x^2+y^2=1$, which meet $x^2+y^2=9$ at A and B. Locus of the point of intersection of tangents at A and B to the cirlce $x^2+y^2=9$ is

• A. $x^2+y^2={\left(\frac{27}{7}\right)}^2$
• B. $x^2-y^2={\left(\frac{27}{7}\right)}^2$
• C. $y^2-x^2={\left(\frac{27}{7}\right)}^2$
• D. None of these

Answer 1

Answer: (A)

$x^2+y^2={\left(\frac{27}{7}\right)}^2$

Since $\Delta$POQ and $\Delta$AOQ are congruent
Hence, $\angle POQ=\angle QOA=\theta$
$cos\theta =\frac{1}{3},$ sing $\angle POR={180}^o$
$\Rightarrow \angle AOR=\pi -2\theta$
Now in triangle AOR, $\angle AOR=\pi -2\theta$ and $AO=3$ unit
$\Rightarrow cos(\pi -2\theta )=\frac{OA}{OR}=\frac{3}{\sqrt{h^2+k^2}}$
$\Rightarrow \sqrt{h^2+k^2}=\frac{27}{7}$
$\Rightarrow x^2+y^2={\left(\frac{27}{7}\right)}^2$