Question

For all real values of $k$, the polar of the point $(2k,k-4)$ with respect to $x^2+y^2-4x-6y+1=0$ passes through the point

• A. $(1,1)$
• B. $(1,-1)$
• C. $(-3,1)$
• D. $(3,1)$

Answer 1

The correct option is D $(3,1)$

Given equation is $x^2+y^2-4x-6y+1=0$

Equation of polar is $T=0$

$\Rightarrow x{x}_1+y{y}_1-4\left(\frac{x+x_1}{2}\right)-6\left(\frac{y+y_1}{2}\right)+1=0\Rightarrow x{x}_1+y{y}_1-2(x+x_1)-3(y+y_1)+1=0\Rightarrow x\left(2k\right)+y(k-4)-2(x+2k)-3(y+k-4)+1=0\Rightarrow 2kx+ky-4y-2x-4k-3y-3k+12=0\Rightarrow k(2x+y-7)-(2x+7y-13)=0$

The equation of polar will be independent of $k$ if its coefficient is zero, if coefficient become zero then rest of the terms will also be equal to zero to satisfy the whole equation.

$2x+y-7=0$ ......(i)

$2x+7y-13$ ........(ii)

Solving (i) and (ii), we get

$\Rightarrow x=3,y=1$

So, the polar passes through the fixed point $(3,1)$.

Hence, option $D$ is correct.