Question

The polar of the circle $x^2+y^2+2λx+2μy+c=0$ with respect to origin will touch $x^2+y^2=r^2,$ if

• A. $r^2=c^2({λ}^2+{μ}^2)$
• B. $c^2=r^2({λ}^2+{μ}^2)$
• C. $c=r({λ}^2+{μ}^2)$
• D. $r=c({λ}^2+{μ}^2)$

Answer 1

Answer: (B)

The equation of polar of the circle
$x^2+y^2+2λx+2μy+c=0$
with respect to origin is
$x.0+y.0+λ(x+0)+μ(y+0)+c=0$
$⇒$ $λx+μy+c=0$
$⇒$ $y=−\frac{λ}{μ}x−\frac{c}{μ}$
If $y=mx+c$ touches the circle $x^2+y^2=r^2,$ then $c=Â\pm \sqrt{1+m^2}$
Hence, $y=−\frac{λ}{μ}x−\frac{c}{μ}$ will touch the circle $x^2+y^2=r^2$
if $−\frac{c}{μ}=Â\pm r\sqrt{1+{\left(−\frac{λ}{μ}\right)}^2}$
$⇒$ $\frac{c^2}{{μ}^2}=r^2\left(\frac{{μ}^2+{λ}^2}{{μ}^2}\right)$
$⇒$ $c^2=r^2({μ}^2+{λ}^2)$