Two circles with centres O and O′ intersect at two points A and B. A line PQ is drawn parallel to OO′ through A(or B) intersecting the circles at P and Q. Prove that PQ = 2 OO′.
Solution:
Given, two circles with centres O and O’ intersect at two points A and B
A line PQ is drawn parallel to OO’ through A or B intersecting the circle at P and Q.
We have to prove that PQ = 2OO’
Join OO’, OP, O’Q, OM and O’N
Considering triangle OPB,
Since OM is the perpendicular bisector of PB
BM = PM ----------------------- (1)
Considering triangle O’BQ,
Since O’N is the perpendicular bisector of BQ
BN = NQ ----------------------- (2)
Adding (1) and (2),
BM + BN = PM + NQ
Adding BM + BN on both sides,
BM + BN + BM + BN = PM + NQ + BM + BN
2(BM + BN) = (BM + PM) + (BN + NQ)
We know, OO’ = MN = BM + BN
So, 2OO’ = BP + BQ
Also, PQ = BP + BQ
2OO’ = PQ
Therefore, PQ = 2OO’
✦ Try This: ABCD is a cyclic quadrilateral in which ∠DBC = 80° and ∠BAC = 40° Find ∠BCD.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 10
NCERT Exemplar Class 9 Maths Exercise 10.3 Problem 18
Two circles with centres O and O′ intersect at two points A and B. A line PQ is drawn parallel to OO′ through A(or B) intersecting the circles at P and Q. Prove that PQ = 2 OO′.
Summary:
Two circles with centres O and O′ intersect at two points A and B. A line PQ is drawn parallel to OO′ through A(or B) intersecting the circles at P and Q. It is proven that PQ = 2 OO′
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