wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

From a point P outside a circle, with centre O, tangents PA and PB are drawn. Then, select the statements that are true.


A

AOP=BOP

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

OP is the bisector of chord AB.

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

AOP=2BOP

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

AOP=7BOP

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

The correct options are
A

AOP=BOP


B

OP is the bisector of chord AB.


Given - A circle with centre O. A point P out side the circle. From P. PA and PB are the tangents to the circle. OP and AB are joined.

In ΔAOP and ΔBOP,

AP = BP (Tangents from P to the circle)

OP = OP (Common)

OA= OB (Radii of the same circle)

Δ AOPΔBOP (ssss postulate)

AOP=BOP (C.P.C.T.)

Now in ΔOAM and ΔOBM,

OA = OB (Radii of the same circle)

OM = OM (Common)

AOM=BOM (Proved AOP=BOP)

ΔOAMΔOBM (S.A.S. Postulate)

AM = MB (C.P.C.T.)

and OMA=OMB (C.P.C.T.)

But OMA+OMB=180 (Linear pair)

OMA=OMB=90

Hence OM or OP is the perpendicular bisector of AB.


flag
Suggest Corrections
thumbs-up
4
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Tangents from External Points
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon