Question

Suppose tangent at a point $P$ on the circle $(x-4{)}^2+y^2=8$ is normal to the circle $x^2+(y-4{)}^2=4.$ Then the square of the distance of this point $P$ from the origin is

• A. $16+24\sqrt{3}$
• B. $32-16\sqrt{3}$
• C. $64+32\sqrt{3}$
• D. $16-8\sqrt{3}$

Answer 1

The correct option is D $16-8\sqrt{3}$

$S_1:(x-4{)}^2+y^2=8\Rightarrow x^2+y^2-8x+8=0$
Let $P(a,b)$ be any point on $S_1.$
Then tangent at $P(a,b)$ to $S_1$ is $T=0$
i.e., $ax+by-\frac{8}{2}(x+a)+8=0$
$\Rightarrow ax+by-4x-4a+8=0$

Above tangent is normal to $x^2+(y-4{)}^2=4,$ so it will pass through $(0,4),$ so we get
$b-a+2=0\dots \left(1\right)$

Also, $P(a,b)$ lies on circle $S_1$
$\Rightarrow a^2+b^2-8a+8=0\dots \left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$, we get
$(a,b)=(3+\sqrt{3},1+\sqrt{3})$
or $(a,b)=(3-\sqrt{3},1-\sqrt{3})$
$\Rightarrow a^2+b^2=16\pm 8\sqrt{3}$