Question

The equation to the locus of the point of intersection of any two perpendicular tangents to $x^2+y^2=4$ is

• A. $x^2+y^2=8$
• B. $x^2+y^2=12$
• C. $x^2+y^2=16$
• D. $x^2+y^2=4\sqrt{3}$

Answer 1

The correct option is A $x^2+y^2=8$

The equation of the tangent to the circle $x^2+y^2=4$ is

$y=mx+2\sqrt{1+m^2}$

$P(h,k)$ lies on the tangent, then

$k-mh=2\sqrt{1+m^2}$

or, $(k-mh{)}^2=4(1+m^2)$

or, $m^2(h^2-4)-2mhk+k^2-4=0$

This is the quadratic equation in $m.$ Let $m_1$ and $m_2$ be roots

$m_1{m}_2=\frac{k^2-4}{h^2-4}=-1$

or, $k^2-4=-h^2+4$

or, $h^2+k^2=8$

Therefore, Equation to the locus of the intersection of any two perpendicular tangents is

$x^2+y^2=8$

Hence, A is the correct option.