Circles Exercise 11.1 |

Circles Exercise 11.2 |

Circles Exercise 11.3 |

**Question 1: A chord of 16 cm is drawn in a circle of radius 10 cm . Find the distance of the chords the centre of the circle.**

**Sol.**

Let AB be a chord of the given circle with center O and radius 10 cm. Then, OA = 10 cm and AB = 16 cm. From O, draw OL ⊥ AB. We know that the perpendicular from the center of a circle to a chord bisects the chord.

∴ \(AL=\frac{1}{2}\times AB\)

= \(=\left ( \frac{1}{2}\times16 \right )\) cm = 8cm

From rt angled Δ OLA, we have

OA ² = OL ² + AL ²

⇒ OL ² = OA ² – AL ²

= 10 ² – 8 ²

= 100 – 64 = 36

∴ \(OL=\sqrt{36}\) = 6 cm

**Question 2: Find the length of a chord which is at a distance of 3 cm from the centre of a circle of radius 5 cm.**

**Sol.**

Let AB be the chord of the given circle with centre O and radius 5 cm.

From O, draw OL ⊥ AB

Then, OA = 5 cm and OL = 3 cm [given]

We know that the perpendicular from the centre of a circle to a chord bisects the chord.

Now, in rt angled Δ OLA, we have

OA ² = OL ² + AL ²

⇒ AL ² = OA ² – OL ²

⇒ AL ² = 5² – 3²

= 25 – 9 = 16

∴ \(AL=\sqrt{16}\) = 4cm

So, AB = 2 AL

= (2 × 4)cm = 8 cm

∴ The length of the chord is 8 cm.

**Question 3: ** **A chord of length 30 cm is drawn at a distance of 8 cm from the centre of the circle. Find out the radius of the circle.**

**Sol.**

Let AB be the chord of the given circle with center O. Draw OL ⊥ AB.

Then, OL is the distance from the center to the chord.

So, we have AB = 30 cm and OL = 8 cm

We know that the perpendicular from the centre of a circle to a chord bisects the chord.

∴ \(AL = \frac{1}{2}\times AB\)

= \(=\left ( \frac{1}{2}\times 30 \right )\) cm = 15 cm

Now, in rt angled Δ OLA, we have

OA ² = OL ² + AL ²

= 8² + 15²

= 64 + 225 = 289

∴ \(OA=\sqrt{289}\) = 17 cm

∴ The radius of the circle is = 17 cm.

**Question 4: In a circle of radius 5 cm, AB and CD are two parallel chords of length 8 cm and 6 CM respectively. Calculate the distance between the chords if they are**

**On the same side of the centre .****On the opposite side of the centre.**

**Sol.**

(i) Let AB and CD be two chords of a circle such that AB || CD which are on the same side of the circle. Also AB = 8 cm and CD = 6 cm, OB = OD = 5 cm. Join OL and LM.

Since the perpendicular from the centre of a circle to a chord bisects the chord.

We have \(LB=\frac{1}{2}\times AB\)

= \(=\left ( \frac{1}{2}\times 8 \right )\) cm = 4 cm

And \(MD=\frac{1}{2}\times CD\)

= \(=\left ( \frac{1}{2}\times 6 \right )\) cm = 3 cm

Now, in rt angled Δ BLO, we have

OB² = LB² + LO²

⇒ LO² = OB² – LB²

= 5² – 4²

= 25 – 16 = 9

∴ \(LO=\sqrt{9}\) = 3 cm

Again in rt angled Δ DMO, we have

OD² = MD² + MO²

MO² = OD² – MD²

= 5² – 3²

= 25 – 9 =16

⇒ \(MO=\sqrt{16}\) = 4 cm

∴ The distance between the chords = (4 – 3) cm = 1 cm.

(ii) Let AB and CD be two chords of a circle such that AB || CD and they are on the oppsite side of the center. AB = 8 cm and CD = 6 cm. Draw OL ⊥ AB and OM ⊥ CD.

Join OA and OC

Then OA = OC + 5 cm (radius)

Since the perpendicular from the centre of a circle to a chord bisects the chord.

We have

\(AL=\frac{1}{2}AB\)= \(\left ( \frac{1}{2}\times 8 \right)\) cm = 4 cm.

Also \(CM=\frac{1}{2}CD\)

= \(\left ( \frac{1}{2}\times 6 \right)\) cm = 3 cm.

Now, in rt angled Δ OLA, we have

OA ² = OL ² + AL ²

⇒ OL ² = OA ² – AL ²

= 5² – 4²

= 25 – 16 = 9 cm

∴ \(OL=\sqrt{9}\) = 3cm

Again in rt angled Δ OMC, we have

OC² = OM² + CM²

⇒ OM² = OC² – CM²

= 5² – 3²

= 25 – 9 = 16

⇒ \(OM=\sqrt{16}\) = 4cm

∴ The distance between the chords = (4 + 3) cm =

**Question 5: In a given figure , the diameter CD of a circle with Centre O is perpendicular to chord Ab . If AB= 12cm and CE= 3 cm .Calculate the radius of the circle.**

**Sol.**

CD is the diameter of a circle with center O, and is perpendicular to chord AB, Join OA.

AB = 12 cm and CE = 3 cm

Let OA = OC = r cm

Then, OE = (r – 3) cm

Since the perpendicular from the centre of a circle to a chord bisects the chord.

We have

\(AE=\frac{1}{2}AB\)= \(\left ( \frac{1}{2}\times 12 \right)\) cm =6 cm

Now, in rt angled Δ OEA, we have

OA ² = OE ² + AE ²

⇒ r² = (r – 3)² + 6²

⇒ = r² -6r + 9 +36

⇒ r² – r² + 6r = 45

⇒ 6r =45

⇒ \(r=\frac{45}{6}\) = 7.5 cm

∴ OA, the radius of the circle is 7.5 cm.

**Question 6: In the given figure, a circle with Centre O is given in which a diameter AB bisects the chord CD at a point E such that CE=ED= 8 cm and EB= 4 cm. Find the radius of the circle.**

**Sol.**

AB is the diameter of a circle with center O which bisects the chord CD at point E.

CE = ED = 8 cm and EB = 4 cm, Join OC.

Let OC = OB = r cm.

Then,

OE = (r – 4) cm

Now, in rt angled Δ OEC, we have

OC ² = OE ² + EC ²

⇒ r² = (r – 4)² + 8²

⇒ r² = r² -8r + 16 +64

⇒ r² – r² + 8r = 80

⇒ 8r =80

⇒ \(r=\frac{80}{8}\) = 10 cm

∴ OE, the radius of the circle is 10 cm.

**Question 7: in the adjoining figure OD is perpendicular to chord AB of a circle with Centre O. If BC is a diameter, show that AC || OD and AC = 2 × OD**

**Sol.**

Given: OD ⊥ AB of a circle with center O, BC is a diameter.

To Prove: AC || OD and AC = 2 × OD

Construction: Join AC.

**Proof: **

We know that the perpendicular from the centre of a circle to a chord bisects the chord.

Here OD ⊥ AB

⇒ D is the mid–point of AB

⇒ AD = BD

Also, O is the mid–point of BC

OC = OB

Now, in Δ ABC, D is the mid – point of AB and O is the mid–point of BC.

Midpoint Theorem: The line segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of it.

∴ OD || AC and \(OD = \frac{1}{2} AC\)

∴ AC = 2 × OD

**Question 8: In the given figure , O is the centre of a circle in which chord AB and CD intersect at PO bisect ∠BPD . Prove that AB = CD .**

**Sol.**

Given: O is the center in which chords AB and CD intersect at P such that bisects ∠BPD

To Prove: AB = CD

Construction: Draw OE ⊥ AB and OF ⊥ CD

**Proof:**

In Δ OEP and Δ OFP

∠ OEP = ∠ OFP [Each equal to 90°]

OP = OP [common]

∠ OPE = ∠ OPF [Since OP bisects ∠ BPD]

Thus, by angle-Side-Angle criterion of congruence, we have

Δ OEP ≅ Δ OFP [By ASA]

The corresponding the parts of the congruent triangles are equal

⇒ OE = OF [CP.CT]

⇒ Chords AB and CD are equidistant from the center O.

⇒ AB = CD [∵ chords equidistant from the center are equal]

∴ AB = CD

**Question 9: Prove that the diameter of the circle perpendicular to one of the two parallel chords of a circle is perpendicular to the other and bisect it .**

**Sol.**

Given: AB and CD are two parallel chords of a circle with center O. POQ is a diameter which is perpendicular to AB.

To Prove: PF ⊥ CD and CF = FD

**Proof: **

AB || CD and POQ is a diameter.

∠ PEB = 90° [Given]

Then, ∠ PFD = ∠ PEB [AB || CD, Corresponding angles]

Thus, PF ⊥ CD

So, OF ⊥ CD

We know that, the perpendicular from the center of a circle to chord, bisect the chord.

∴ CF = FD.

**Question 10: Prove that two different cycle cannot intersect each other at more than two points.**

**Sol.**

If possible let two different circles intersect at three distinct point A,B and C.

Then, these points are non collinear. Do a unique circles can be drawn to pass through these points. This is a contradiction.

**Question 11: Two circle of radius 10 cm and 8 cm intersect each other, and the length of the common chord is 12 Cm. Find the distance between the centres.**

**Sol.**

OA = 10 cm and AB = 12 cm

∴ \(AD=\frac{1}{2}AB\)

= \(\left ( \frac{1}{2}\times 12 \right)\) cm =6 cm

Now, in rt angled Δ ADO, we have

OA ² = AD ² + OD ²

⇒ OD ² = OA ² – AD ²

= 10² – 6²

= 100 – 36 = 64

⇒ \(OD=\sqrt{64}\) = 8 cm

Again, we have O’A = 8 cm

In rt angled Δ ADO’, we have

O’A ² = AD ² + O’D ²

⇒ O’D ² = O’A ² – AD ²

= 8² – 6²

= 64 – 36 = 28

⇒ \(OD=\sqrt{28}\) = \(2\sqrt{7}\) cm

∴ O’O = (OD + O’D)

= \((8+2\sqrt{7})\) cm

∴ The distance between their centers is \((8+2\sqrt{7})\) cm

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