If the line `l x+m y+n=0` touches the circle `x^2+y^2=a^2` , then prove that `(l^2+m^2)^2=n^2dot`

Options

A. `((a^(2)l)/(n), (a^(2)m)/(n))`
B. `(a(-a^(2)l)/(n), (-a^(2)m)/(n))`
C. `((-a^(2)n)/(n), (-a^(2)n)/(m))`
D. none of these

Answer

B

Explanation

Let the pole be `(x_(1), y_(1))`. Then, the equation of the polar is
`x x_(1)+y y_(2)=a^(2) " " ...(i)`
Since (i) and lx+my+n=0 represent the same straight line.
`:. (x_(1))/(l)=(y_(1))/(m)=(-a^(2))/(n) rArr x_(1)=(-a^(2)l)/(n), y_(1)=(-a^(2)m)/(n)`
Hence, the pole is `((-a^(2)l)/(n), (-a^(2)m)/(n))`