Question

A circle with centre at (2, 4) is such that the line $x+y+2=0$ cuts a chord of length 6. Find the radius of the circle.

• A. $\sqrt{41}$
• B. $\sqrt{11}$
• C. $\sqrt{21}$
• D. $\sqrt{31}$

Answer 1

The correct option is A $\sqrt{41}$

The length of perpendicular from $(2,4)$ on the line $x+y+2=0$ would be $\frac{2+4+2}{\sqrt{2}}=\frac{8}{\sqrt{2}}=4\sqrt{2}$

The perpendicular from the centre onto a chord bisects it.

So, the base of the triangle so formed would be $3$.

The radius being the hypotenuse of the right angled triangle
Radius $=\sqrt{3^2+(4\sqrt{2}{)}^2}=\sqrt{9+32}=\sqrt{41}$