Question

Tangents drawn from the origin to the circle $x^2+y^2-2px-2qy+q^2=0$ are perpendicular to each other if

• A. $p^2=q^2$
• B. $p^2-q^2=1$
• C. $p^2+q^2=1$
• D. None of these

Answer 1

Answer: (A)

$p^2=q^2$

The equation of pair of tangents drawn from the origin to the given circle are $S{S}_1=T^2$
$\Rightarrow (x^2+y^2-2px-2qy+g^2)(0+0-0-0+g^2)={(x.0+y.0-p(x+0)-q(y+0)+y^2)}^2$
$\Rightarrow q^2(x^2+y^2-2px-2qy+g^2)-{(-px-qy+g^2)}^2=0$
The two tangents are $\perp$ if $g^2+q^2-p^2-g^2=0$
(Sum of coefficient of $x^2+y^2=0$)
$\Rightarrow q^2=p^2$