Question

Two circles touch each other internally at a point P. A chord AB of the bigger circle intersects the other circle in C and D. Then, $\angle CPA=$ _______.

• A. $5\angle DPB$.
• B. $2\angle DPB$.
• C. $8\angle DPB$.
• D. $\angle DPB$.

Answer 1

The correct option is D

$\angle DPB$.

Given - Two circles touch each other internally at P. A chord AB of the bigger circle, intersects the smaller circle at C and D, AP, BP, CP and DP are joined.

Construction - Draw a tangent TS at P to the circles given.

$\because TPS$ is the tangent, PD is the chord.

$\therefore \angle PAB=\angle BPS$ ....(i)

(Angles in alternate segment)

Similarly we can prove that

$\angle PCD=\angle DPS$ .....(ii)

Subtracting (i) and (ii), we get

$\angle PCD-\angle PAB=\angle DPS-\angle BPS$

But in $\Delta PAC$

Ext. $\angle PCD=\angle PAB+\angle CPA$

$\therefore \angle PAB+\angle CPA-\angle PAB=\angle DPS-\angle BPS$

$\angle CPA=\angle DPB$