Question

Let $C$ be the circle with centre at $\left(1,1\right)$ and radius $1$. If $T$ is the circle centred at $\left(0,k\right)$ passing through the origin and touching the circle $C$ externally, then the radius of $T$ is equal to

• A. $\frac{\sqrt{3}}{\sqrt{2}}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is D

$\frac{1}{4}$

Explanation for correct option:

Finding the radius of $T$:

Plotting the graph using the given information.

The distance formula is given by $AB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

$\begin{array}{rcl}\therefore C_1{C}_2& =& \sqrt{{\left(0-1\right)}^2+{\left(k-1\right)}^2}\\ & =& \sqrt{1+{\left(k-1\right)}^2}\dots \dots \left(1\right)\end{array}$

We know $$\begin{gathered} r_1=1,r_2=k \\ {r}_1+r_2=C_1{C}_2\dots \dots \left(2\right) \end{gathered}$$

Equating Equation (1) and (2) we get

$1+y=\sqrt{1+{\left(1-y\right)}^2}$

Squaring on both side

$$\begin{gathered} \Rightarrow {\left(1+k\right)}^2=1+{\left(1-k\right)}^2 \\ \Rightarrow 1+k^2+2k=1+1-2k+k^2 \\ \Rightarrow 2k+2k=1 \\ \Rightarrow 4k=1 \\ \Rightarrow k=\frac{1}{4} \end{gathered}$$

Therefore, option (D) is the correct answer.