SSC CPO 2020 PYQ — Question 131

The correct answer is Option 3 i.e. 6 + √161.

Let O and O' be the centers of the circle of radii 15 cm and 10 cm respectively.

Let AB be their common chord.

We have,

OA = 15 cm, O'A = 10 cm, and AB = 16 cm

So, AP = (1/2) × AB

⇒ 16/2 = 8 cm

In the right triangle OPA, we have

OA² = OP² + PA²

⇒ OA = $\sqrt{OP^2 - PA^2}$

⇒ OA = $\sqrt{15^2-8^2}$

OA = $\sqrt161$ cm

In the right triangle O'PA we have

O'A² = AP² + O'P²

⇒ O'A = $\sqrt{10^2-8^2}$

⇒ O'A = $\sqrt36$

⇒ O'A = 6 cm

⇒ OO' = OP + O'P

⇒ OO' = 6 + $\sqrt161$

Thus,

The distance between centers is  6 + $\sqrt161$ cm