Question

If the locus of the mid-point of the line segment from the point $(3,2)$ to a point on the circle, $x^2+y^2=1$ is a circle of the radius $r$, then $r$ is equal to

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

Answer 1

The correct option is B

$\frac{1}{2}$

Explanation for the correct option:

Step 1: Derive a relation between the general point on the given circle and the mid-point.

In the question, an equation of the circle $x^2+y^2=1$ and a point $(3,2)$ is given.

We know that the standard equation of a circle is given by ${(x-x_1)}^2+{(y-y_1)}^2=r^2$, where $(x,y)$ is the general point on the circle, the centre of the circle is at $(x_1,y_1)$ and $r$ is the radius of the circle.

Assume that, the mid-point of the line segment from the point $(3,2)$ to a point on the circle, $x^2+y^2=1$ is $(h,k)$.

So, $\left(\frac{x+3}{2},\frac{y+2}{2}\right)=(h,k)$

Now, compute $x$ and $y$ is terms of $h$ and $k$ respectively.

Therefore,

$$\begin{gathered} \frac{x+3}{2}=h \\ \Rightarrow x=2h-3 \end{gathered}$$

And

$$\begin{gathered} \frac{y+2}{2}=k \\ \Rightarrow y=2k-2 \end{gathered}$$

Therefore, $(x,y)=\left(2h-3,2k-2\right)$.

Step 2: Drive an equation for the locus of the mid-point.

Since, $(x,y)$ lies on the given circle. So, it satisfy the given equation of the circle.

So,

$$\begin{gathered} {\left(2h-3\right)}^2+{\left(2k-2\right)}^2=1 \\ \Rightarrow \frac{{\left(2h-3\right)}^2}{4}+\frac{{\left(2k-2\right)}^2}{4}=\frac{1}{4} \\ \Rightarrow {\left(h-\frac{3}{2}\right)}^2+{\left(k-1\right)}^2={\left(\frac{1}{2}\right)}^2 \end{gathered}$$

Which represents the locus of the mid-point of the line segment from the point $(3,2)$ to a point on the circle, $x^2+y^2=1$.

Since, the radius of the locus of the mid-point of the line segment from the point $(3,2)$ to a point on the circle, $x^2+y^2=1$ is $\frac{1}{2}$.

Therefore, the value of $r$ is $\frac{1}{2}$.

Hence, option (B) $\frac{1}{2}$ is the correct answer.