Two circles touch each other externally at point P in the given figure. AB is the direct common tangent of these circles. Then ÐAPB equals to:

60°

No worries! We‘ve got your back. Try BYJU‘S free classes today!

90°

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

75°

No worries! We‘ve got your back. Try BYJU‘S free classes today!

120°

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
90°

AM = MP [tangents from an external point]
$∠ M A P$ = $∠ M P A$
Similarly, MP = MB
$∠ M B P$ = $∠ M P B$
Now in $∠ A B P$
$∠ P A B$ + $∠ A B P$ + $∠ A P B$ = $180^{\circ}$
$∠ M P A$ + $∠ M P B$ + $∠ M P A$ + $∠ M P B$ = $180^{\circ}$
$∠ M P A$ + $∠ M P B$ = $90^{\circ}$
So, $∠ A P B$ = $90^{\circ}$

Suggest Corrections

Similar questions

In the given figure, two circles touch each other externally at point $P$ . $A B$ is the direct common tangent of these circles. Prove that :

(i) tangent at point $P$ bisects $A B$ ,

(ii) angle $A P B =$ $90^{\circ}$ .

In the given figure, two circles touch each other externally at point, P. AB is the direct common tangent of these circles. Select the statements that are true.

Q. Two equal circles touch each other externally at C and AB is a common tangent to the circles . Then, ∠ACB =

(a) 60°

(b) 45°

(c) 30°

(d) 90°

Q. Let two circles touch each other externally and

P

is the point of intersection of there direct common tangents. If the direct common tangents from

P

touches the circles at

A

and

B

on same side and

P A = A B = 4

, then the radius of smaller circle (in units) is

Q. Two circles touch each other externally at C and AB is common tangent to the circles then

∠ A C B

is :

Join BYJU'S Learning Program

Two circles touch each other externally at point P in the given figure. AB is the direct common tangent of these circles. Then ÐAPB equals to:

The correct option is B 90° AM = MP [tangents from an external point] ∠MAP = ∠MPA Similarly, MP = MB ∠MBP = ∠MPB Now in ∠ABP ∠PAB + ∠ABP + ∠APB =180∘ ∠MPA +∠MPB +∠MPA +∠MPB = 180∘ ∠MPA +∠MPB = 90∘ So, ∠APB = 90∘