NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.3
NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.3 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Chapter 10 Circles Ex 10.3.
Ex 10.3 Class 9 Maths Question 1.
Draw different pairs of circles.
How many points does each pair have in common? What is the maximum number of
common points?
Solution.
Different pairs of circles are given below:
(i) If they intersect each other, two
points are common.
Hence,
the maximum number of common points is two.
Ex 10.3 Class 9 Maths Question 2.
Suppose you are given a circle. Give a construction
to find its centre.
Solution.
- Take three points P, Q and R on the
circle.
- Join PQ and QR.
- Draw MT and NS, which are respectively the
perpendicular bisectors of PQ and QR and intersecting each other at a
point O.
Hence,
O is the centre of the circle.
Ex 10.3 Class 9 Maths Question 3.
If two circles intersect at two points, prove that
their centres lie on the perpendicular bisector of the common chord.
Solution:
We have two circles with centres O
and O’, intersecting at A and B.
∴ AB is the common chord of two
circles and OO’ is the line segment joining their centres.
Let OO’ and AB intersect each other at M.
OA = OB [Radii of the same circle]
O’A = O’B [Radii of the same circle]
OO’ = OO’ [Common]
∴ ∆OAO’ ≅ ∆OBO’ [By SSS congruence rule]
⇒ ∠1 = ∠2 [C.P.C.T.]
Now, in ∆AOM and ∆BOM, we have
OA = OB [Radii of the same circle]
OM = OM [Common]
∠1 = ∠2 [Proved above]
∴ ∆AOM = ∆BOM [By SAS congruence rule]
⇒ ∠3 = ∠4 [C.P.C.T.]
But ∠3 + ∠4 = 180° [Linear pair]
∴ ∠3 = ∠4 = 90°
⇒ AM ⊥ OO’
Also, AM = BM [C.P.C.T.]
⇒ M is the mid-point of AB.
Thus, OO’ is the perpendicular bisector of AB.
NCERT Solutions for Maths Class 10