**Circles**

**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:**

Let us draw different pairs of circles as shown below:

We have

Figure | Maximum number of common points |

(i) | nil |

(ii) | one |

(«i) | two |

Thus, two circles can have at the most two points in common.

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

**Solution:**

Steps of construction :**Step I :** Take any three points on the given circle. Let these points be A, B and C.**Step II : **Join AB and BC.**Step III :** Draw the perpendicular bisector, PQ of AB.**Step IV:** Draw the perpendicular bisector, RS of BC such that it intersects PQ at O.

Thus, ‘O’ is the required centre of the given drcle.

**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.

∴ To prove that OO’ is the perpendicular bisector of AB,

we join OA, OB, O’A and O’B. Now, in ∆QAO’ and ∆OBO’,

we have

OA = OB [Radii of the same circle]

O’A = O’B [Radii of the same circle]

OO’ = OO’ [Common]

∴ ∆OAO’ ≅ ∆OBO’ [By SSS congruence criteria]

⇒ ∠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 criteria]

⇒ ∠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.