Question

If the pole of the polar w.r.t the circle $x^2+y^2=c^2$ is the circle $x^2+y^2=9c^2$. Then, this polar will be the tangent of the circle.

• A. $x^2+y^2=\frac{9}{4}{c}^2$
• B. $x^2+y^2=\frac{c^2}{9}$
• C. $x^2+y^2=4c^2$
• D. $Noneofthese$

Answer 1

The correct option is B $x^2+y^2=\frac{c^2}{9}$

Given, circle eq'n is $x^2+y^2=c^2$

Center of the circle is $(0,0)$

let the equation of the polar be $l_{1}x+m_{1}y+n_1=0$

Foot of the perpendicular A' from origin to the line $l_{1}x+m_{1}y+n_1=0$ is $A^{\prime }(\frac{-n_1{l}_1}{l_1^2+m_1^2},\frac{-n_1{m}_1}{l_1^2+m_1^2})$

Distance of A' from center is OA'$=\sqrt{\frac{n_1^2}{l_1^2+m_1^2}}$

Let the pole of this line be A.

Then , $OA.O{A}^{\prime }=c^2$

$O{A}^2.O{A}^{\prime 2}=c^4$

$O{A}^2=\frac{c^4(l_1^2+m_1^2)}{n_1^2}$

Let $A=(x,y)$

Then, $x^2+y^2=\frac{c^4(l_1^2+m_1^2)}{n_1^2}...\left(1\right)$

$A,O,A^{\prime }$ are collinear.

Therefore, $\frac{x}{y}=\frac{l_1}{m_1}$

$y=\frac{m_1}{l_1}x$

substituting in (1) ,

$x^2=\frac{c^4{l}_1^2}{n_1^2},x=\pm \frac{c^2{l}_1}{n_1},y=\pm \frac{c^2{m}_1}{n_1}$

given, pole is on the circle $x^2+y^2=9c^2$

So, $c^2(l_1^2+m_1^2)=9n_1^2$

let the circle eq'n be $x^2+y^2=k^2$

Tangent to this circle will be of the form $y=mx+k\sqrt{1+m^2}$

$m=\frac{-l_1}{m_1}$

$k\sqrt{1+m^2}=\frac{-n_1}{m_1}$

$k^2(1+m^2)=\frac{n_1^2}{m_1^2}$

$k^2(1+\frac{l_1^2}{m_1^2})=\frac{n_1^2}{m_1^2}$

$k^2(l_1^2+m_1^2)=n_1^2$

$k^2\frac{9n_1^2}{c^2}=n_1^2$

$k^2=\frac{c^2}{9}$

eq'n of the circle is $x^2+y^2=\frac{c^2}{9}$