Question

Two circles of radii $10$ cm and $8$ cm intersects each other and the length of the common chord is $12$ cm, find the distance between their centers.

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

Answer 1

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

Given length of common chord $AB=12$ cm

Let the radius of the circle with centre $O$ is $OA=10$ cm

Radius of circle with centre $P$ is $AP=8$ cm

From the figure, $OP\perp AB$

$\Rightarrow AC=CB$

$\therefore AC=6$ cm (Since $AB=12$ cm)

In $\Delta ACP$,

$A{P}^2=P{C}^2+A{C}^2$ [By Pythagoras theorem]

$\Rightarrow 8^2=P{C}^2+6^2$

$\Rightarrow P{C}^2=64–36=28$

$PC=2\sqrt{7}$ cm

Consider $\Delta ACO$,

$A{O}^2=O{C}^2+A{C}^2$ [By Pythagoras theorem]

$\Rightarrow {10}^2=O{C}^2+6^2$

$\Rightarrow O{C}^2=100-36=64$

$\Rightarrow OC=8$ cm

From the figure, $OP=OC+PC=8+2\sqrt{7}$ cm

Hence, the distance between the centres is $(8+2\sqrt{7})$ cm.