Question

The tangents drawn from the origin to the circle $x^2+y^2-2kx-2ry+r^2=0$ are perpendicular, if:

• A. $k=r$
• B. $k=-r$
• C. $k^2=r^2$
• D. $k^2+r^2=1$

Answer 1

Answer: (A), (B), (C)

$k=r$
$k^2=r^2$
$k=-r$

Given circle equation is $x^2+y^2-2kx-2ry+r^2=0$.
Pair of tangents drawn from an external point $(x_1,y_1)$ to $S=0$ is given by $S{S}_{11}=S_1^2$
$\Rightarrow (x^2+y^2-2kx-2ry+r^2)r^2={(-kx-ry+r^2)}^2$
above equation represents a pair of perpendicular lines if sum of coefficient of $x^2$ and $y^2$ is $0$.
$\Rightarrow r^2+r^2=k^2+r^2$
$\therefore k^2=r^2$
Hence, options A,B and C.