Question

Two circles with centres O and O' intersect at points A and B. A line PQ is drawn to O O' through A (or B) intersecting the circles at P and Q. Prove that PQ = 2OO'

Answer 1

Answer:

Step-by-step explanation:

Given: There are two circles with centres O and O' that intersect at points A and B.

To prove: PQ=2OO'

Construction: Join P and Q which is parallel to OO', Join OO' an draw OR and O'S which being the perpendicular bisectors.

Proof: In ΔOPB,

PR=RB (OR is the perpendicular bisector of PB)

and in ΔO'BQ,

BS=SQ(O'S is the perpendicular bisector of BQ)

Now, BR+BS=PR+SQ

Adding, BR+BS on both ides, we have

2(BR+BS)=BR+BS+PR+SQ

2(BR+BS)=(BR+PR)+(BS+SQ)

2(BR+BS)=BP+BQ

2(BR+BS)=PQ (Because BP+BQ+PQ)

Now, BR+BR=RS

2(RS)=PQ

As OO'=RS, therefore

2OO'=PQ

Hence proved.

Answer 2

Answer:

Step-by-step explanation:

Given: There are two circles with centres O and O' that intersect at points A and B.

To prove: PQ=2OO'

Construction: Join P and Q which is parallel to OO', Join OO' an draw OR and O'S which being the perpendicular bisectors.

Proof: In ΔOPB,

PR=RB (OR is the perpendicular bisector of PB)

and in ΔO'BQ,

BS=SQ(O'S is the perpendicular bisector of BQ)

Now, BR+BS=PR+SQ

Adding, BR+BS on both ides, we have

2(BR+BS)=BR+BS+PR+SQ

2(BR+BS)=(BR+PR)+(BS+SQ)

2(BR+BS)=BP+BQ

2(BR+BS)=PQ (Because BP+BQ+PQ)

Now, BR+BR=RS

2(RS)=PQ

As OO'=RS, therefore

2OO'=PQ

Hence proved.