# Class 9 NCERT Solutions – Chapter 10 Circles – Exercise 10.3

### 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**:

(i) Two points commonAttention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the

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(ii) One point common

(iii) One point common

(iv) No point common

(v) No point commonAs we can analyse from above, two circles can cut each other maximum at two points.

### Question 2. Suppose you are given a circle. Give a construction to find its centre.

**Solution:**

Let the circle be C1

We need to find its centre.

Step 1:Take points P, Q, R on the circle

Step 2:Join PR and RQ.We know that perpendicular bisector of a chord passes through centre

So, we construct perpendicular bisectors of PR and RQ

Step 3:Take a compass. With point P as pointy end and R as pencil end of the compass, mark an arc above and below PR. Do same with R as pointy end P as pencil end of the compass.

Step 4:Join points intersected by the arcs.The line formed is the perpendicular bisector of PR.

Step 5:Take compass, with point R as pointy end and Q as pencil end of the compass mark an arc above and below RQ.Do the same with Q as pointy end and R as pencil end of the compass

Step 6:Join the points intersected by the arcs.The line formed is the perpendicular bisector of RQ.

Step 7:The point where two perpendicular bisectors intersect is the centre of the circle. Mark it as point O.Thus, O is the centre of the given circle.

### Question 3: If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

**Solution:**

Given,

Let circle C1 have centre O and circle C2 have centre X, PQ is th common chord.

To prove:OX is the perpendicular bisector of PQ i.e.1. PR = RQ

2. ∠PRO = ∠PRX = ∠QRO = ∠QRX = 90°

Construction:Join PO, PX, QO, QX

Proof:In △POX and △QOX

OP = OQ (Radius of circle C1)

XP = XQ (Radius of circle C2)

OX = OX (Common)

∴ △POX ≅ △QOX (SSS Congruence rule)

∠POX = ∠QOX (CPCT) —-(1)

Also,

In △POR and △QOR

OP = OQ (Radius of circle C1)

∠POR = ∠QOR ( From (1))

OR = OR (Common)

∴ △OPX ≅ △OQX (SAS Congruence Rule)

PR = QR (CPCT)

& ∠PRO = ∠QRO (CPCT) —-(2)

Since PQ is a line

∠PRO + ∠QRO = 180° (Linear Pair)

∠PRO + ∠PRO= 180° ( From (2))

2∠PRO = 180°

∠PRO = 180° / 2

∠PRO = 90°

Therefore,

∠QRO = ∠PRO = 90°

Also,

∠PRX = ∠QRO = 90° (Vertically opposite angles)

∠QRX = ∠PRO = 90° (Vertically opposite angles)

Since, ∠PRO = ∠PRX = ∠QRO = ∠QRX = 90°

∴ OX is the perpendicular bisector of PQ