Question

With respect to the circle $x^2+y^2+6x-8y-10=0$

• A. The chord of contact of tangents from $(2,1)$ is $5x-3y-8=0$
• B. The pole of the line $5x-3y-8=0$ is $(2,1)$
• C. The polar of the point $(2,1)$ is $5x-3y-8=0$
• D. All of these

Answer 1

Answer: (B), (C)

The correct options are
B The pole of the line $5x-3y-8=0$ is $(2,1)$
C The polar of the point $(2,1)$ is $5x-3y-8=0$

Given circle is $S\equiv x^2+y^2+6x-8y-9=0$

Since $S_{(2,1)}$ $=4+1+12-8-10=-1<0,$

So the point $(2,1)$ lies inside the given circle.

$\therefore$ Chord of contact of tangents from $(2,1)$does not exist.

Also, polar of $(2,1)$ w.r.t. the circle is

$2x+y+3(x+2)-4(y+1)-10=0$

$\Rightarrow 5x-3y-8=0.$