Question

The line $3x-2y=k$ meets the circle $x^2+y^2=4r^2$ at only one point, if $k^2$ is

• A. $20r^2$
• B. $52r^2$
• C. $\frac{52}{9}{r}^2$
• D. $\frac{20}{9}{r}^2$

Answer 1

The correct option is B $52r^2$

Since the line $3x-2y=k$ is a tangent to the circle $x^2+y^2=4r^2$,

perpendicular distance of the line from the centre of the circle must equal the radius.

$\therefore \frac{|k|}{\sqrt{3^2+2^2}}=2r$

$\therefore |k|=2r\sqrt{13}$

$\therefore k^2=52r^2$