AP State Syllabus AP Board 9th Class Maths Solutions Chapter 12 Circles Ex 12.3 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 12th Lesson Circles Exercise 12.3

Question 1.

Draw the following triangles and construct circumcircles for them.

(OR)

In ΔABC, AB = 6 cm, BC = 7 cm and ∠A = 60°.

Construct a circumcircle to the triangle XYZ given XY = 6cm, YZ = 7cm and ∠Y = 60°. Also, write steps of construction.

Solution:

Steps of construction :

- Draw the triangle with given mea-sures.
- Draw perpendicular bisectors to the sides.
- The point of concurrence of per-pendicular bisectors be S’.
- With centre S; SA as radius, draw a circle which also passes through B and C.
- This is the required circumcircle.

ii) In ΔPQR; PQ = 5 cm, QR = 6 cm and RP = 8.2 cm.

Steps of construction:

- Draw ΔPQR with given measures.
- Draw perpendicular to PQ, QR and RS; let they meet at ‘S’.
- With S as centre and SP as radius draw a circle.
- This is the required circumcircle.

iii) In ΔXYZ, XY = 4.8 cm, ∠X = 60°and ∠Y = 70°.

Steps of construction:

- Draw ΔXYZ with given measures.
- Draw perpendicular bisectors to the sides of ΔXYZ, let the point of con-currence be S’.
- Draw the circle (S, \(\overline{\mathrm{SX}}\)).
- This is the required circumcircle.

Question 2.

Draw two circles passing A, B where AB = 5.4 cm.

(OR)

Draw a line segment AB with 5.4 cm. length and draw two different circles that passes through both A and B.

Solution:

Steps of construction:

- Draw a line segment AB = 5.4 cm.
- Draw the perpendicular bisector \(\stackrel{\leftrightarrow}{X Y}\) of AB.
- Take any point P on \(\stackrel{\leftrightarrow}{X Y}\).
- With P as centre and PA as radius draw a circle.
- Let Q be another point on XY.
- Draw the circle (Q, \(\overline{\mathrm{QA}}\)).

Question 3.

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

Solution:

Let two circles with centre P and Q intersect at two distinct points say A and B.

Join A, B to form the common chord

\(\overline{\mathrm{AB}}\). Let ‘O’ be the midpoint of AB.

Join ‘O’ with P and Q.

Now in ΔAPO and ΔBPO

AP = BP (radii)

PO = PO (common)

AO = BO (∵ O is the midpoint)

∴ ΔAPO ≅ ΔBPO (S.S.S. congruence)

Also ∠AOP = ∠BOP (CPCT)

But these are linear pair of angles.

∴ ∠AOP = ∠BOP = 90°

Similarly in ΔAOQ and ΔBOQ

AQ = BQ (radii)

AO = BO (∵ O is the midpoint of AB)

OQ = OQ (common)

∴ AAOQ ≅ ABOQ

Also ∠AOQ = ∠BOQ (CPCT)

Also ∠AOQ + ∠BOQ = 180° (linear pair of angles)

∴ ∠AOQ = ∠BOQ = \(\frac{180^{\circ}}{2}\) = 90°

Now ∠AOP + ∠AOQ = 180°

∴ PQ is a line.

Hence the proof.

Question 4.

If two intersecting chords of a circle make equal angles with diameter pass¬ing through their point of intersection, prove that the chords are equal.

Solution:

Let ‘O’ be the centre of the circle.

PQ is a drametre.

\(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) are two chords meeting at E, a point on the diameter.

∠AEO = ∠DEO

Drop two perpendiculars OL and OM from ‘O’ to AB and CD;

Now in ΔLEO and ΔMEO

∠LEO = ∠MEO [given]

EO = EO [Common]

∠ELO = ∠EMO [construction 90°]

∴ ΔLEO ≅ ΔMEO

[ ∵ A.A.S. congruence]

∴ OL = OM [CPCT]

i.e., The two chords \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) are at equidistant from the centre ‘O’.

∴ AB = CD

[∵ Chords which are equi-distant from the centre are equal]

Hence proved.

Question 5.

In the given figure, AB is a chord of circle with centre ‘O’. CD is the diam-eter perpendicular to AB. Show that AD = BD.

Solution:

CD is diameter, O is the centre.

CD ⊥ AB; Let M be the point of inter-section.

Now in ΔAMD and ΔBMD

AM = BM [ ∵ radius perpendicular to a chord bisects it]

∠AMD =∠BMD [given]

DM = DM (common)

∴ ΔAMD ≅ ΔBMD

⇒ AD = BD [C.P.C.T]