**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.

**(ii)**If they touch each other externally or internally, only one point is common.

**(iii)**If they do not intersect or touch each other, no points are common.

From figures, it is obvious that these pairs may have 0 or 1 or 2 points in common. Hence, a pair of circles cannot intersect each other at more than two points.

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.**

**Steps of Construction:**

- 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**