**Telangana**** SCERT Solution Class IX (9) Math Chapter 12 Circles Exercise 12.3**

**(1.) Draw the following triangles and construct circumcircles for them. **

**(i) In Δ ABC, AB = 6cm, BC = 7 cm and ****∠****A = 60 ^{o} **

**(ii) In Δ PQR, PQ = 5cm, QR = 6 cm and RP = 8.2cm **

**(iii) In Δ XYZ, XY = 4.8cm, ****∠****X = 60 ^{o} and **

**∠**

**Y = 70**

^{o}**Solution:**

(i)

In ABC, AB = 6cm, BC = 7cm, ∠A = 60^{o}

Step 1: – Draw line AB = 6cm

Step 2: – Draw a line 60^{o} from point A on line AB

Step 3: – Cut the line at form B taking radius 7cm on compass. Hence, BC = 7cm

Step 4: – Join AC and BC at point C to from △ABC

Step 5: – Draw perpendicular bisectors of AB and AC from base AB and AC

Step 6: – Point where they intersects is the centre of circle

Step 7: – Draw a circle taking radius OA which interact at point B and C

**(ii) **

In △PQR, PQ= 5cm, QR = 6Cm and RP = 8.2cm

Step 1:- Draw a line PQ = 5cm

Step 2: – Take radius 8.2 cm on compass and draw an arc from point P

Step 3:- Take radius 6cm on compass draw an arc that cuts the previous arc from point Q

Step 4: – Joint the intersection point R at the arc with point P & Q to get △PQR of side 5cm, 8.2cm, & 6cm

Step 5: – Draw, perpendicular bisects for PQ & QR using compass of any radius from each ends of line segments PQ and QR

Step 6: – The intersection of as bisects is the centre of circle O

Step 7: – Draw a circle of radius OP from centre O connecting points R and Q

**(iii)**

In △XYZ, XY = 4.8Cm, ∠X = 60^{o} & ∠Y = 70^{o}

Step 1: – Draw One segment XY = 4.8cm

Step 2: – Draw a line at 60^{o} from point X at Line segment XY

Step 3: – Draw a line at 70^{o} from point Y at Line segment XY

Step 4: – Draw perpendicular bisectors of line XY and YZ by drawing arc from each end of of XY and YZ taking any radius

Step 5: – The point where the perpendicular bisects cuts each other is the centre of circle O

Step 6: – Draw a circle of radius OX connecting point Z & Y

**(2.) Draw two circles passing through A, B where AB = 5.4 cm**

**Solution:**

Step 1: – Draw a line AB = 5.4 cm

Step 2: – Draw perpendicular bisects of line AB by drawing arcs of any radius from each ends of AB on opposite sides of AB

Step 3: – The point where the arcs into cuts each other arc the centre of circles P & Q

Step 4: – Draw a circle taking radius PA with P as centre

Step 5: – Draw a circle taking radius QA at centre Q

∴ AB being the chord of two circle.

**(3.) If two circles intersect at two points, then prove that their centres lie on the perpendicular bisector of the common chord.**

**Solution:**

From the above demonstration we can see that the centres of circle lie on the same perpendicular bisects of chord AB.

**(4.) If two intersecting chords of a circle make equal angles with diameter passing through their point of intersection, prove that the chords are equal.**

**Solution:**

Given,

PQ diameter

AB, CD are two chords intersecting each other at E on diameter PQ.

OL & OM are perpendiculars from centre of circle O to chords AB & CD respectively

∠AEQ = ∠DEQ [given]

∴ In △EOL & △EON

(i) ∠LED = ∠NEO [∠AEQ = ∠DEQ, similar angle]

(ii) ∠ELO = ∠ENO = 90^{o} [⊥ was at chord AB & CD]

(iii) EO is common side

∴ △EOL ≅ △EOM by AAS

∠O = MO [∵ corresponding sides of congruent triangles are equal]

∴ AB = CD [∵∠ O = MO which are perpendicular of AB & CD & chords equidistance from the centre of circle are equal in length]

**(5) In the adjacent figure, AB is a chord of circle with centre O. CD is the diameter perpendicualr to AB. Show that AD = BD.**

**Solution:**

**Solution:**

Given, CD diameter perpendicular to chord AB. O being the centre of circle

Let, the point where the diameter & Chord intersect be P.

∴In △APD & △BPD

(i) DP is the common side

(ii) ∠BPD = ∠APD = 90^{o} [diameter DC ⊥ to chord AB]

(iii) AP = BP [∵perpendicular drawn from the centre of circle O bisect chord AB at point P]

△APD ≅ △BPD by SAS

AD = BD [∵corresponding sides of congruent triangle are equal in length]