Question

The line $2x-y+1=0$ is tangent to the circle at the point $(2,5)$ and the centre of the circles lies on $x-2y=4$. The radius of the circle is

• A. $3\sqrt{5}$
• B. $5\sqrt{3}$
• C. $2\sqrt{5}$
• D. $5\sqrt{2}$

Answer 1

Answer: (A)

$3\sqrt{5}$

Given:

Tangent, $T:2x-y+1=0$

Slope of tangent, $T$ to the circle is $2$

The line joining the centre and the point of contact of tangent is perpendicular with the tangent.

Thus, slope of line $OA=-\frac{1}{2}$

Equation of $OA$,

$(y-5)=-\frac{1}{2}(x-2)$

$x+2y=12$

$\therefore$ Intersection with $x-2y=4$ will give coordinates of centre which are $(8,2)$

$\therefore r=OA=\sqrt{(8-2{)}^2+(2-5{)}^2}=3\sqrt{5}$

Hence, option A.