**Chapter 3 Circle Set 3.1**

## Chapter 3 Circle Set 3.1

**Question 1.In the adjoining figure, the radius of a circle with centre C is 6 cm, line AB is a tangent at A. Answer the following questions.i. What is the measure of ∠CAB? Why?ii. What is the distance of point C from line AB? Why?iii. d(A, B) = 6 cm, find d(B, C).iv. What is the measure of ∠ABC? Why?**

Solution:

Solution:

i. line AB is the tangent to the circle with centre C and radius AC. [Given]

∴ ∠CAB = 90° (i) [Tangent theorem]

ii. seg CA ⊥ line AB [From (i)]

radius = l(AC) = 6 cm

∴ The distance of point C from line AB is 6 cm.

iii. In ∆CAB, ∠CAB = 90° [From (i)]

∴ BC

^{2}= AB

^{2}+ AC

^{2}. [Pythagoras theorem]

= 6

^{2}+ 6

^{2}

= 2 × 6

^{2}

∴ d(B, C) = 6 cm

iv. In ∆ABC,

AC = AB = 6cm

∴ ∠ABC = ∠ACB [Isosceles triangle theorem]

Let ∠ABC = ∠ACB =x

In ∆ABC,

∠CAB + ∠ABC + ∠ACB = 180° [Sum of the measures of angles of a triangle is 180°]

∴ 90° + x + x = 180°

∴ 90 + 2x = 180°

∴ 2x = 180°- 90°

∴ x = 90

^{0}/2

∴ x = 45°

∴ ∠ABC = 45°

**Question 2.In the adjoining figure, O is the centre of the circle. From point R, seg RM and seg RN are tangent segments touching the circle at M and N. If (OR) = 10 cm and radius of the circle = 5 cm, theni. What is the length of each tangent segment?ii. What is the measure of ∠MRO?iii. What is the measure of ∠MRN?Solution:**

seg RM and seg RN are tangents to the circle with centre O. [Given]

∴ ∠OMR = ∠ONR = 90° [Tangent theorem]

i. In ∆OMR, ∠OMR = 90°

∴ OR

^{2}= OM

^{2}+ RM

^{2}[Pythagoras theorem]

∴ 10

^{2}= 5

^{2}+ RM

^{2}

∴ 100 = 25 + RM

^{2}

∴ RM

^{2}= 75

∴ RM = RN [Tangent segment theorem]

Length of each tangent segment is 5√3 cm.

ii. In ∆RMO,

∠OMR = 90° [Tangent theorem]

OM = 5 cm and OR = 10 cm

∴ OM = 1/2 OR

∴ ∠MRO = 30° (i) [Converse of 30° – 60° – 90° theorem]

Similarly, ∠NRO = 30°

iii. But, ∠MRN = ∠MRO + ∠NRO [Angle addition property]

= 30° + 30° [From (i)]

∴ ∠MRN = 60°

**Question 3.Seg RM and seg RN are tangent segments of a circle with centre O. Prove that seg OR bisects ∠MRN as well as ∠MON.Solution:**

Proof:

In ∆OMR and ∆ONR,

seg RM ≅ seg RN [Tangent segment theorem]

seg OM ≅ seg ON [Radii of the same circle]

seg OR ≅ seg OR [Common side]

∴ ∆OMR ≅ ∆ONR [SSS test of congruency]

{∴ ∠MRO ≅ ∠NRO

∠MOR ≅ ∠NOR } [c.a.c.t.]

∴ seg OR bisects ∠MRN and ∠MON.

**Question 4.What is the distance between two parallel tangents of a circle having radius 4.5 cm? Justify your answer.Solution:**

Let the lines PQ and RS be the two parallel tangents to circle at M and N respectively. Through centre O, draw line AB || line RS.

OM = ON = 4.5 [Given]

line AB || line RS [Construction]

line PQ || line RS [Given]

∴ line AB || line PQ || line RS

Now, ∠OMP = ∠ONR = 90° (i) [Tangent theorem]

For line PQ || line AB,

∠OMP = ∠AON = 90° (ii) [Corresponding angles and from (i)]

For line RS || line AB,

∠ONR = ∠AOM = 90° (iii) [Corresponding angles and from (i)]

∠AON + ∠AOM = 90° + 90° [From (ii) and (iii)]

∴ ∠AON + ∠AOM = 180°

∴ ∠AON and ∠AOM form a linear pair.

∴ ray OM and ray ON are opposite rays.

∴ Points M, O, N are collinear. (iv)

∴ MN = OM + ON [M – O – N, From (iv)]

∴ MN = 4.5 + 4.5

∴ MN = 9 cm

∴ Distance between two parallel tangents PQ and RS is 9 cm.

**Question 1.In the adjoining figure, seg QR is a chord of the circle with centre O. P is the midpoint of the chord QR. If QR = 24, OP = 10, find radius of the circle. To find solution of the problem, write the theorems that are useful. Using them, solve the problem. (Textbook pg. no. 48)Solution:**

Theorems which are useful to find solution:

i. The segment joining the centre of a circle and the midpoint of a chord is perpendicular to the chord.

ii. In a right angled triangle, sum of the squares of the perpendicular sides is equal to square of its hypotenuse.

QP = 1/2 (QR) [P is the midpoint of chord QR]

1/2 × 24 = 12 units

Also, seg OP ⊥ chord QR [The segment joining centre of a circle and midpoint of a chord is perpendicular to the chord]

In ∆OPQ, ∠OPQ = 90°

∴ OQ

^{2}= OP

^{2}+ QP

^{2}[Pythagoras theorem]

= 10

^{2}+ 12

^{2}

= 100 + 144

= 244

**Question 2.In the adjoining figure, M is the centre of the circle and seg AB is a diameter, seg MS ⊥ chord AD, seg MT ⊥ chord AC, ∠DAB ≅ ∠CAB.i. Prove that: chord AD ≅ chord AC.ii. To solve this problem which theorems will you use?a. The chords which are equidistant from the centre are equal in length.b. Congruent chords of a circle are equidistant from the centre.iii. Which of the following tests of congruence of triangles will be useful?a. SAS b. ASA c. SSS d. AAS e. Hypotenuse-side test.Using appropriate test and theorem write the proof of the above example. (Textbook pg. no, 48)Solution:**

Proof:

i. ∠DAB ≅ ∠CAB [Given]

∴ ∠SAM ≅ ∠TAM (i) [A – S – D, A – M – B, A -T – C]

In ∆SAM and ∆TAM,

∠SAM ≅ ∠TAM [From (i)]

∠ASM ≅ ∠ATM [Each angle is of measure 90°]

seg AM ≅ seg AM [Common side]

∴ ∆SAM ≅ ∆TAM [AAS [SAA] test of congruency]

∴ side MS ≅ side MT [c.s.c.t]

But, seg MS ⊥ chord AD [Given]

seg MT ⊥chord AC

∴ chord AD ≅ chord AC [Chords of a circle equidistant from the centre are congruent]

ii. Theorem used for solving the problem:

The chords which are equidistant from the centre are equal in length.

iii. Test of congruency useful in solving the above problem is AAS ISAAI test of congruency.

**Question 3.i. Draw segment AB. Draw perpendicular bisector l of the segment AB. Take point P on the line l as centre,PA as radius and draw a circle. Observe that the circle passes through point B also. Find the reason.ii. Taking any other point Q on the line l, if a circle is drawn with centre Q and radius QA, will it pass through B? Think.iii. How many such circles can be drawn, passing through A and B? Where will their centres lie? (Textbook pg. no. 49)Solution:**

i. Draw the circle with centre P and radius PA.

line l is the perpendicular bisector of seg AB.

Every point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment.

∴ PA = PB … [Perpendicular bisector theorem]

∴ PA = PB = radius

∴ The circle with centre P and radius PA passes through point B.

ii. The circle with any other point Q and radius QA is drawn.

QA = QB = radius … [Perpendicular bisector theorem]

∴ The circle with centre Q and radius QA passes through point B.

iii. We can draw infinite number of circles passing through A and B.

All their centres will lie on the perpendicular bisector of AB (i.e., line l)

**Question 4.i. Take any three non-collinear points. What should be done to draw a circle passing through all these points? Draw a circle through these points.ii. Is it possible to draw one more circle passing through these three points? Think of it. (Textbook pg. no. 49)Solution:**

i. Let points A, B, C be any three non collinear points.

Draw the perpendicular bisector of seg AB (line l).

∴ Points A and B are equidistant from any point of line l ….(i)[Perpendicular bisector theorem]

Draw the perpendicular bisector of seg BC (line m) to intersect line l at point P.

∴ Points B and C are equidistant from any point of line m ….(ii) [Perpendicular bisector theorem]

∴ PA = PB …[From (i)]

PB = PC … [From (ii)]

∴ PA = PB = PC = radius

∴ With PA as radius the required circle is drawn through points A, B, C.

ii. It is not possible to draw more than one circle passing through these three points.

**Question 5.Take 3 collinear points D, E, F. Try to draw a circle passing through these points. If you are not able to draw a circle, think of the reason. (Textbook pg. no. 49)Solution:**

Let D, E, F be the collinear points.

The perpendicular bisector of DE and EF drawn (i.e., line l and line m) do not intersect at a common point.

∴ There is no single common point (centre of circle) from which a circle can be drawn passing through points D, E and F.

Hence, we cannot draw a circle passing through points D, E and F.

**Question 6.Which theorem do we use in proving that hypotenuse is the longest side of a right angled triangle? (Textbook pg. no. 52)Solution:**

In ∆ABC,

∠ABC = 90°

∴ ∠BAC < 90° and ∠ACB < 90° [Given]

∴ ∠ABC > ∠BAC and ∠ABC > ∠ACB

∴ AC > BC and AC > AB [Side opposite to greater angle is greater]

∴ Hypotenuse is the longest side in right angled triangle.

We use theorem, If two angles of a triangle are not equal, then the side opposite to the greater angle is greater than the side opposite to the smaller angle.

**Question 7.Theorem: Tangent segments drawn from an external point to a circle are congruentDraw radius AP and radius AQ and complete the following proof of the theorem.Given: A is the centre of the circle.Tangents through external point D touch the circle at the points P and Q.To prove: seg DP ≅ seg DQConstruction: Draw seg AP and seg AQ.Solution:**

Proof:

In ∆PAD and ∆QAD,

seg PA ≅ [segQA] [Radii of the same circle]

seg AD ≅ seg AD [Common side]

∠APD = ∠AQD = 90° [Tangent theorem]

∴ ∆PAD = ∆QAD [By Hypotenuse side test]

∴ seg DP = seg DQ [c.s.c.t]