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.

solution

Answer 2

Appropriate 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 OO' || CBC' and OO' = ½ CBC'.

$\large\underline{\sf{Solution-}}$

Given that,

• 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.

• AB is perpendicular to CBC'.

As it is given that, AB is perpendicular to CBC'.

So, ∠ABC = 90°

It implies AC is diameter of circle.

So, O lies on AC

Similarly, O' lies on AC'

As AC is diameter, so O is the midpoint of AC

Also, AC' is a diameter, so O' is the midpoint of AC'

Now, In triangle ACC'

$\rm \: O \: is \: the \: midpoint \: of \: AC$

and

$\rm \: O' \: is \: the \: midpoint \: of \: AC'$

We know,

Midpoint Theorem states that line segment joining the midpoints of two sides of a triangle is parallel to third side and equals to half of it.

So, Using Midpoint Theorem, we have

$\rm\implies \:OO' \: \parallel \: CBC'$

and

$\rm\implies \:OO' \: = \: \dfrac{1}{2} \: CBC'$

Hence, Proved

$\rule{190pt}{2pt}$

Additional information :-

1. Angle in semi-circle is right angle.

2. Angle in same segments are equal.

3. Angle subtended at the centre of a circle by an arc is double the angle subtended on the circumference of a circle by the same arc.

4. Sum of the opposite angles of a cyclic quadrilateral is supplementary.