Question

Two circles with center o and o' intersect at two points P and Q such that OP and O'Q are tangents to two circles. Find the length of the chord PQ

Answer 1

Answer:

 $PQ=\frac{2PO.PO'}{OO'}$

Step-by-step explanation:

in figure OP is perpendicular to O'P and OQ is perpendicular to O'Q as radius at point of contact is perpendicular to the tangent

OP=O'P

OQ=O'Q [tangent fron same external point are equal]

by SAS congruence ⇒ ΔOO'P≅ΔOO'Q

In ΔOO'P

∠O'PA=∠O'QA [∵O'P=O'Q]

from CPCT⇒∠PO'A=∠QO'A

In ΔOO'P

∠O'+∠P+∠Q=180° [Angle sum property]

2∠P+∠O'=180°

2∠P+2∠PO'A=180°

∠P+∠PO'A=190°

∴∠PAO'=90° [By applying angle sum in ΔPAO']

So ∠PAO=90° [Linear pair]

inΔOO'P and ΔOPA

sin∠POA=PA/PO=PO'/OO'

PA=PO.PO'/OO'

MULTIPLY 2 ON BOTH SIDES

2PA=2(PO.PO')/OO'

 $PQ=\frac{2PO.PO'}{OO'}$ [radius⊥ chord bisects it so 2PA=PQ]