Question

The radius of the circle, which is touched by the line$y=x$ and has its center on the positive direction of the $x-$axis and also cuts off a chord of length$2$ units along the line $\sqrt{3}y-x=0$ is

• A. $\sqrt{5}$
• B. $\sqrt{3}$
• C. $\sqrt{2}$
• D. $1$

Answer 1

The correct option is C

$\sqrt{2}$

Explanation for the correct option:

Solve for the radius of the circle.

Let $\left(a,0\right)$be the center of the circle with radius $r$

$y-x=0$ is tangent to the circle.

So, distance of the centre from the tangent is equal to radius.

$$\begin{gathered} r=\frac{\left|o-a\right|}{\sqrt{1^2+1^2}} \\ \Rightarrow a^2=2r^2...\left(1\right) \end{gathered}$$

Now perpendicular distance of the given chord is $=\frac{\left|0-{\displaystyle \frac{a}{\sqrt{3}}}\right|}{\sqrt{{\displaystyle \frac{1}{3}}+1}}=\frac{a}{2}$

$$\begin{gathered} \therefore C{A}^2=r^2=C{N}^2+N{A}^2 \\ \Rightarrow r^2=\frac{a^2}{4}+1^2 \\ \Rightarrow r^2=\frac{2r^2}{4}+1^2 \\ \Rightarrow \frac{r^2}{2}=1 \\ \Rightarrow r^2=2 \\ \Rightarrow r=\sqrt{2} \end{gathered}$$

Hence, option(C) is the correct answer.