Question

If line $x+2y=1$ is the polar of circle $x^2+y^2=5$, then pole is

• A. $(5,10)$
• B. $(5,5)$
• C. $(-5,10)$
• D. $(-5,-10)$

Answer 1

Answer: (A), (D)

The correct options are
A $(5,10)$
D $(-5,-10)$

Given, circle eq'n is $x^2+y^2=a^2=5$

Given line eq'n is $x+2y-1=0$

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

Foot of the perpendicular 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 , $OA.O{A}^{\prime }=a^2$

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

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

Let $A=(x,y)$

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

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

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

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

substituting in (1) ,

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

substituting the values of $l_1,m_1$,

$l_1=1,m_1=2,n_1=-1$

$x=\pm 5,y=\pm 10$