Question

Locus of the point of intersection of perpendicular tangents to the circle $x^2+y^2=16$ is

• A. $x^2+y^2=8$
• B. $x^2+y^2=32$
• C. $x^2+y^2=64$
• D. $x^2+y^2=16$

Answer 1

The correct option is B

$x^2+y^2=32$

Explanation for the correct option:

Given the equation of the circle

$$\begin{gathered} x^2+y^2=16 \\ \Rightarrow x^2+y^2=4^2 \end{gathered}$$

To find the locus of the point of intersection of perpendicular tangent.

We know that ,

Radius of locus = $\sqrt{r^2+r^2}=\sqrt{2r^2}$

From the given equation radius is given as $4$

Put $r=4$ in $\sqrt{2r^2}$

$\Rightarrow$Radius of locus $=\sqrt{2\times 4^2}=\sqrt{32}$

Therefore, the equation of the tangent

$$\begin{gathered} \Rightarrow x^2+y^2={\left(\sqrt{32}\right)}^2 \\ \Rightarrow x^2+y^2=32 \end{gathered}$$

Hence, option (B) is the correct answer.