Question

Two Circles with centres O and O' intersect at two points A and B. Two chords are drawn through A to intersect the circles at points C and C, respectively, such that CBC is a straight line and AB is perpendicular to CBC. Prove that 00'= CBC. Prove that OO' = ½ CBC'.​

Answer 1

Answer:

Firstly draw two circles with center O and O’ such that they intersect at A and B.

Draw a line PQ parallel to OO’.

In the circle with center O, we have:

OP and OB are the radii of the circle. PB is the chord with OM as its perpendicular bisector.

i.e. BM=MP....(1)

In the circle with center O’, we have:

O’B and O’Q are the radii of the circle. BQ is the chord with O’N as its perpendicular bisector.

i.e. BN=NQ....(1)BM=MP....(1)

From (1) and (2), we have:

BM+BN=MP+NQ

⇒(BM+BN)+(BM+BN)=(BM+BN)+(MP+NQ)

⇒2(BM+BN)=(BM+BN)+(MP+NQ)

⇒2(OO’)=(BM+MP)+(BN+NQ)

⇒2(OO’)=BP+BQ

⇒2OO’=PQ

Hence, proved.

Step-by-step explanation:

Hope its helpful

Answer 2

Step-by-step explanation:

Construct OM⊥PQ and O

′

N⊥PQ

So we get

OM⊥AP

We know that the perpendicular from the centre of a circle bisects the chords

We know that

O

′

Q

We know that the perpendicular frmo the centre of a circle bisects the chord

AN=QN

It can be written as

AQ=2AN...(2)

So we get

PQ=AP+PQ

By substituting the values

PQ=2(AM+AN)

We get

PQ=2MN

From the figure we know that MNO

′

O is a rectangle

PQ=2OO

′

Therefore, it is proved that PQ=2OO

′