Question

If two chords having lengths ${\mathrm{a}}^2-1$ and $3\left(\mathrm{a}+1\right)$, where $\mathrm{a}$ is a constant of a circle bisect each other, then the radius of the circle is

• A. $6$
• B. $\frac{15}{2}$
• C. $8$
• D. $\frac{19}{2}$
• E. $10$

Answer 1

The correct option is B

$\frac{15}{2}$

Explanation for the correct answer:

Finding the radius of the circle:

Two chords of a circle bisect each other only when both are the diameter of a circle.

As both the chords are diameter,

$$\begin{gathered} {\mathrm{a}}^2-1=3\left(\mathrm{a}+1\right) \\ \Rightarrow {\mathrm{a}}^2-3\mathrm{a}-4=0 \\ \Rightarrow {\mathrm{a}}^2-4\mathrm{a}+\mathrm{a}-4=0 \\ \Rightarrow \mathrm{a}\left(\mathrm{a}-4\right)+1\left(\mathrm{a}-4\right)=0 \\ \Rightarrow \left(\mathrm{a}-4\right)\left(\mathrm{a}+1\right)=0 \\ \Rightarrow \mathrm{a}=4\mathrm{or}-1 \end{gathered}$$

But radius cannot be negative

So, $\mathrm{a}=4$ and the radius of the circle is,

$\begin{array}{rcl}& =& \frac{3\left(\mathrm{a}+1\right)}{2}\left\{\because \mathrm{radius}=\frac{\mathrm{d}}{2}\right\}\\ & =& \frac{15}{2}\end{array}$

Hence, the correct option is B.