Question

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

The radius of the circle is

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

Answer 1

The correct option is A

$\sqrt{41}$

The explanation for the correct option:

Step1. Draw the diagram :

The length of the perpendicular from $(2,4)$ on the line $x+y+2=0$ is

$AC=\frac{\left|2\right(1)+4(1)+2|}{\sqrt{1^2+1^2}}=\frac{8}{\sqrt{2}}=4\sqrt{2}$

Step2. To find the radius of the circle:

Let $r$ be the radius of the circle.

The perpendicular from the center onto a chord bisects it.

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

$AB=3$

In a right-angled triangle $△CAB$,

$$\begin{gathered} r^2={\left(AC\right)}^2+{\left(AB\right)}^2 \\ \Rightarrow r^2={\left(4\sqrt{2}\right)}^2+{\left(3\right)}^2 \\ \Rightarrow r^2=32+9 \\ \Rightarrow r=\sqrt{41} \end{gathered}$$

Hence the correct option is (A).