Question

in the given figure,there are two circles intersecting each other at A and B.prove that the line joining their centres bisects the common chord AB

Answer 1

Hey there,

GIVEN: Circles with center O & O' , AC is tangent to the circle with center o' at A. OO' meets AB at D.

TO PROVE THAT: angle BAO = angle CAO PROOF: In triangle O'AO

O'A = O'O ( being radii of the same circle with center O')

So, angle O'AO = angle O'OA = y……..(1)

O'A is perpendicular to tangent AC ( as theorem states that tangent to any circle at any point is perpendicular to the radius through that point.)

So, Angle O’AC = 90°…………..(2)

Eq(2) - Eq(1)

Angle O'AC — AngleO'AO

= Angle OAC = 90°— y ………..(3)

Now since ,

O is equidistant from A& B ( being radii of the same circle with center O)

That implies that O lies on the perpendicular bisector of the segment AB.

Similarly, O' is equidistant from A & B( being radii of the same circle with center O')

That implies that O' lies on the perpendicular bisector of the segment AB.

Hence conclude that OO' is perpendicular bisector of AB , at D ( by theorem )

Now, in Triangle ADO

Angle ADO = 90° ( proved above)

Angle DOA = y ( by eq (1) )

Therefore, third angle OAD =90° — y………(4)

By comparing eq(3) & (4)

We have, angle OAC = angle OAD

[Hence proved]

Hope this helps!

Answer 2

Answer:

GIVEN :

2 circles with centre P and Q and having a common chord A and B

TO PROVE : AO=BO

PROOF :

In triangle AOP and BOP

\_AOP=\_BOP

(If 2 circles intersect at 2 points, their centres lie on the perpendicular bisector of the common chord)

OP=OP (common)

AP=BP (radii of circle)

Therefore, ∆AOP ≈ ∆BOP (by RHS congruency)

AO=BO (by CPCT)

$<b>HENCE PROVED$