Question

Tangents drawn from $(2,0)$ to the circle $x^2+y^2=1$ touches the circle $A$ and $B$. Then

• A. $A=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right),B=(-\frac{1}{2},-\frac{\sqrt{3}}{2})$
• B. $A=(-\frac{1}{2},\frac{-\sqrt{3}}{2}),B=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$
• C. $A=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right),B=\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$
• D. $A=\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right),B=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$

Answer 1

Answer: (C), (D)

The correct options are
C $A=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right),B=\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$
D $A=\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right),B=\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$

Since, tangents from $(2,0)$ intersect the circle at A and B

Then AB is the polar of the point $(2,0)$ w.r.t circle $x^2+y^2=1$

equation of polar with pole $(2,0)$ is $S_1=x{x}_1+y{y}_1-1=0$

$\Rightarrow 2x+0\left(y\right)=1$
$\Rightarrow x=\frac{1}{2}$
Substituting in $x^2+y^2=1$
we get $y=\pm \frac{\sqrt{3}}{2}$