Question

Two circles of radii $10cm$ and $8cm$ intersect each other and the length of the common chord is $12m$. Then the distance between their centres is

• A. $(10+2\sqrt{7})cm$
• B. $(8+2\sqrt{7})cm$
• C. $(12+2\sqrt{7})cm$
• D. $(6+2\sqrt{7})cm$

Answer 1

The correct option is B $(8+2\sqrt{7})cm$

We know that the line joining the centres of two circles bisects the common chord at right angles.

$\therefore MN\perp PQ,\angle MRP={90}^o=\angle MRQ$

And $RM=\frac{1}{2}MN=\frac{1}{2}\times 12cm=6cm$

So, $\Delta MRP$ is a right angled triangle with $PM$ as hypotenuse.

$\therefore$ Applying Pythagoras theorem, we have

${PM}^2={PR}^2+{RM}^2$

$\Rightarrow {10}^2={PR}^2+6^2$

$\Rightarrow PR=8cm$

Similarly,

$\Delta MRQ$ is a right angled triangle with $QM$ as hypotenuse.

$\therefore$ Applying Pythagoras theorem, we have

${QM}^2={RQ}^2+{RM}^2$

$\Rightarrow 8^2={RQ}^2+6^2$

$\Rightarrow RQ=2\sqrt{7}cm$

$\therefore PQ=PR+RQ=(8+2\sqrt{7})cm$

Hence, option B.