## Solid Geometry Chapter 1 Theorems Related To Circle

**What is circle and its different parts**

Definition of a circle, different parts of a circle, such as centre, radius, diameter, circumference, chord, etc. What are concentric and congruent circles, you have already learnt it.

Some properties and characteristics of a circle have also been discussed there. So, we shall discuss here no more about them.

In this chapter, we shall at first discuss about some properties and theorems related to a circle.

You know that the straight line joining any two points on the circumference of a circle is called a chord of the circle. If that straight line passes through the centre of the circle, then it is called the diameter of the circle.

**WBBSE Solutions for Class 10 Maths**

Diameter is the greatest chord of a circle and the number of diameters of a circle is infinity. Any chord of a circle divides it in two parts.

If the two parts be equal, then each part is called a semi-circle Whereas, if the two parts are not equal, then the greater part is called Major segment and the smaller part is known as Minor segment.

If the centres of some tildes be the same, then they are called concentric circles. The radii of the concentric circles are not equal.

However, if instead of different centres, the radii of two or more than two circles be equal, then they are known as Equal or congruent circles.

**Wbbse Class 10 Maths Solutions**

Any chord of a circle subtends an angle at the centre of the circle.

Such as, in the following figures, the chord AB together with two of its radii OA and OB have subtended an angle ∠AOB at the centre of the circle, chord PQ together with two radii PX and QX of the circle have subtended an angle ∠PXQ at the centre of the circle and the chord RS along with two of its radii YR and YS have subtended an angle ∠RYS at the centre of the circle.

Now if the lengths of the chords be equal, i.e., if AB = PQ = RS, then ∠AOB = ∠PXQ – ∠RVS, i.e., the subtended angles are equal.

So, we can say that the chords of the circle, making equal subtended angles at the centre arc equal in length.

It is true conversely.

**Wbbse Class 10 Maths Solutions**

**To determine the centre of a circle **

Let ABCD be a given circle. We have to determine the centre of this circle. Then at first, we construct the chords AB and CD. Then let us construct the perpendicular bisector PM and QN of AB and CD respectively.

Let PM and QN intersect each other at O. Then O is the centre of that circle.

Thus by drawing perpendicular bisectors of any two chords of a circle we can determine the centre of the circle.

**To draw a circle passing through three non-collinear points and theorem related to it **

Let A, B and C are three non-collinear points. We have to draw a circle passing through these three points.

Since the circle will pass through A, B and C, the centre of the circle must be equidistant from A, B and C.

Now to locate a point equidistant from A, B and C, we first draw the line segments AB and BC.

**Wbbse Class 10 Maths Solutions**

Then let us draw the perpendicular bisectors of AB and BC. Let MN and PQ be two perpendicular bisectors of AB and CD respectively and MN and PQ intersect at O.

Then point O will be equidistant from points A, B and C, i.e. O is the centre of the circle.

Now, taking O as the centre and OA or OB or OC as the radius, let us draw a circle. Then that circle will be the required circle.

Now the question is, is it possible to draw an other circle passing through these three points? In reply we can say that one and only one circle can be drawn through these three non-collinear points.

Since, the non-collinear points A, B and C are fixed, So both AB and BC are also fixed.

As a result the perpendicular bisectors of AB and BC are also fixed and which assures that the. centre O is also a fixed point. That is why the radius OA is fixed.

**Wbbse Class 10 Maths Solutions**

So, one and only one circle can be drawn through three non-collinear fixed points.

We shall now prove this theorem with perfect reason geometrically.

## Solid Geometry Chapter 1 Theorems Related To Circle Theorems

**Theorem 1. Prove that only one circle can be drawn through three non-collinear points.**

** Given**: Let A, B and C be three non-collinear points.

** To prove**: We shall have to prove that, one and only one circle can be drawn through these three non-collinear points.

** Construction**: Let us join points A, B and B, C.

Thus we get two line segments AB and BC.

Now, let us draw perpendicular bisectors PQ of AB and RS of BC respectively.

Let PQ and RS intersect each other at point O.

PQ and RS intersect the line segments AB and BC at M and N respectively.

Let us join O, A, O, B and O, C and taking O as the centre, OA, OB or OC as the radius, let us draw a circle. Then this circle is our required circle.

** Proof**: In ΔOAM and ΔOBM, AM = BM

[PM is the perpendicular bisector of AB.]

∠OMA = ∠OMB [each is rightangle] and OM is common to both.

∴ ΔOAM ≅ ΔOBM

OA = OB [corresponding sides of congruent triangles]

Similarly, it can be proved that OB =’ OC.

∴ OA = OB = OC

∴ the circle drawn by taking O as the centre and OA as the radius must passes through the points A, B and C which are non-collinear.

Hence one and only one circle can be drawn through three non-collinear points. (Proved)

We have seen in above that the centre of a circle lies on the perpendicular bisector of any chord of the circle.

Conversely, if perpendiculars are drawn to any chord (except the diameter) of a circle, then that perpendicular bisects the chord.

We shall now prove this logically by the process of geometry.

**Theorem 2. The perpendicular drawn to any chord (except the diameter) from the centre of a circle bisects the chord.**

** Given**: Let AB is a chord (except a diameter) and OP ⊥ AB when OP intersects AB at P.

** To prove**: We shall have to prove that OP bisects AB, i.e., AP = PB.

** Construction **•: Let us join O, A and O, B.

** Proof**: OP ⊥ AB, both ΔOAP and ΔOBP are right-angled.

Now in ΔOAP and ΔOBP, ∠OPA = ∠OPB, [each is rightangle]

hypotenuse OA = hypotenuse OB [radii of same circle] and OP is common to both.

∴ ΔOPA = ΔOPB [As per R – H – S congruency]

∴ AP = PB [corresponding sides of congruent triangles]

∴ AP = PB (Proved)

We shall now prove the converse theorem logically.

**Theorem 3. (Converse theorem of theorem 2) Prove that if any straight line passing through the centre of a circle bisects any chord, which is not a diameter, then the straight line will be perpendicular on that chord.**

**Given:** Let PQ is a chord of the circle with the centre at O and M is the middle point of PQ, i.e., PM = QM.

**To prove:** We shall have to prove that OM ⊥ PQ, i.e., OM is perpendicular to the chord PQ.

**Construction:** Let us join O, P and O, Q.

**Proof:** In ΔOPM and ΔOQM, OP = OQ [radius of the same circle]

PM = QM [M is the middle point of PQ.] and OM is common to both.

∴ ΔOPM ≅ OQM [By the condition of S – S – S congruency of triangles]

∴ ∠OMP = ∠OMQ [similar angles of two congruent triangles]

But they are two adjacent angles too, obtained when OM stands upon PQ and they are equal to each other.

∴ each angle is a right angle, i.e., ∠OMP = ∠OMQ = 1 right angle.

∴ OM ⊥ PQ. (Proved)

In the following examples how the above theorems have been applied in real problems is discussed thoroughly.

## Solid Geometry Chapter 1 Theorems Related To Circle Multiple Choice Questions

**Example 1. AB is a diameter of a circle with centre O, and AC = CB; the value of ∠CAB is**

- 40°
- 45°
- 50°
- 55°

Solution: ∠ACB = 90° [Semi Circle angle]

In ΔABC, AC = CB

∠CAB = ∠ABC = \(\frac{180^{\circ}-90^{\circ}}{2}\) = 45°

∴ The correct answer is 2. 45°

The value of ∠CAB is 2. 45°.

**Example 2. POR diameter of a circle. If PR = 12 cm, QR = 16 cm, then the length of PQ is**

- 10 cm
- 20 cm
- 30 cm
- 40 cm

Solution: In ΔPQR. ∠PRQ = 90° [Semi Circular angle]

PQ^{2} = PR^{2} + QR^{2} [From Pythagoras theorem]

= (12° + 16^{2}) cm^{2} = (144 + 256) cm^{2}

= 400 cm^{2}

PQ = √400 = 20 cm

∴ The correct answer is 2. 20 cm

The length of PQ is 2. 20 cm

**Example 3. The lengths of two chords AB and CD of the circle with centre at O are equal. If ∠AOB = 60°, then the value of ∠COD is**

- 40°
- 50°
- 60°
- 70°

Solution: We know that equal chords of a circle subtend equal angles at the centre. the chords AB and CD also subtend equal angles at the centre.

∠AOB = ∠COD or, 60° = ∠COD.

Hence. ∠COD = 60°

∴ 3. 60° is correct.

**The value of ∠COD is 3. 60°**

**Example 4. The radius of a circle is 13 cm and one of its chord is 10 cm. Then the distance of the chord from the centre is**

- 12 cm
- 14 cm
- 16 cm
- 20 cm

Solution: Let the length of the chord AB is 10 cm.

O is the centre of the circle and OC ⊥ AB.

∴ the required distance = OC.

As per the question, the radius of the circle, OA =13 cm,

Now, from the right angled triangle OAC, we get,

OA^{2} = OC^{2} + AC^{2} [OA = hypotenuse]…….. (1) [by Pythagoras theorem]

Again, OC ⊥ AB, ∴C is the mid-point of AB.

AC= \(\frac{1}{2}\) AB = \(\frac{1}{2}\) x 10 cm = 5 cm

∴ from (1) we get, 13^{2} = OC^{2} + 5^{2} [OA = 13 cm]

or, 169 = OC^{2} + 25

or, OC^{2} =169-25

or, OC^{2} = 144

⇒ OC = √144 =12

Hence the required distance = 12 cm

∴ 1. 12 is correct.

** The distance of the chord from the centre is 1. 12**

**Example 5. If O is the centre of circle and QR is the diameter then the value of ****x ****is**

- 45°
- 55°
- 60°
- 70°

** Solution**: In ΔPOQ, OP = OQ [radii of same circle]

∴ ∠OQP = ∠OPR = 55°

i.e. ∠PQR = 55°

∠PSR = ∠PQR [angles in the same segment]

x° = 55°

∴ The correct answer is 2. 55°

**Class 10 Maths Wbbse Solutions**

**Example 6. O is the centre of circle, ∠BAC = x°, ∠AOC = 70° and ∠AOB = 130°, then the value of x is**

- 60°
- 80°
- 100°
- None of these

Solution: ∠BOC = 360° – ∠AOB – ∠AOC

= 360° – 130° – 70° = 160°

∠BAC = \(\frac{1}{2}\) ∠BOC ; x°= \(\frac{1}{2}\) x 160°, x° = 80°

∴ The correct answer is 2. 80°

**The value of ****x ****is 2. 80°**

**Example 7.** **The centre of two concentric circles is O. A straight line intersects one of the circles at A and B and the other circle at C and D respectively. If AC = 5 cm, then the length of BD is**

- 2.5 cm
- 5 cm
- 3 cm
- 6 cm

Solution: Let AB intersects the greater one of two concentric circles with centre at O at points A and B and the smaller one at C and D.

∴ Let OP ⊥ CD, OP ⊥ AB.

∴ P is the mid-point of both CD and AB.

∴ PC = PD and PA – PB

Now, BD = PB – PD

= PA – PC [PB = PA and PD = PC] = AC

= 5 cm [AC = 5 cm given]

∴ length of BD = 5cm

∴ 2. 5 cm is correct.

** The length of BD is 2. 5 cm**

**Class 10 Maths Wbbse Solutions**

## Solid Geometry Chapter 1 Theorems Related To Circle True Or False

**Example 1. The lengths of the radii of two congruent circles are equal.**

Solution: The statement is true.

**Example 2. The arc is a line segment.**

Solution: The statement is false.

**Example 3. PS and QT are the perpendiculars on side QR and RP of the triangle PQR. P, Q, S, T are concyclic.**

Solutions: ∠PTQ = ∠PSQ = 90° as a line segment PQ joining two points P and Q subtends equal angles at two other points S and T on the same of PQ.

∴ P, Q, S, T points are concyclic.

∴ The statement is true.

**Class 10 Maths Wbbse Solutions**

**Example 4. The opposite angle of a cyclic quadrilateral is complimentary.**

Solution: The statement is false.

## Solid Geometry Chapter 1 Theorems Related To Circle Fill In The Blanks

**Example 1. All angles in the same segments are ______**

Solution: Equal

**Example 2. The vertics of square is _____**

Solution: Concyclic.

**Class 10 Maths Wbbse Solutions**

**Example 3. If the ratio of two chords AB and CD of a circle with its centre O is 1 : 1, then ∠AOB : ∠COD = ______**

Solution: Ratio of length of chord AB and CD of a circle with its centre O is 1 : 1.

∴ AB = CD

∠AOB = ∠COD

∠AOB : ∠COD =1:1

∴ 1:1.

## Solid Geometry Chapter 1 Theorems Related To Circle Short Answer Type Questions

**Example 1. Two equal circles of radius 13 cm intersect each other and the length of their common chord is 10 cm. Find the distance between the centres of the circles.**

Solution:

**Given:**

Two equal circles of radius 13 cm intersect each other and the length of their common chord is 10 cm.

Let A and B be two centres of the circle and PQ is their common chord.

Let the distance between the centres be AB which intersects PQ at O.

∴ O is the mid-point of PQ [AO ⊥ PQ]

∴ PO = \(\frac{1}{2}\) PQ = \(\frac{1}{2}\) x 10 cm = 5 cm. [PQ = 10 cm]

According to the question, PA = 13 cm.

Now, in the right-angled triangle AOP, PA^{2} = OP^{2} + OA^{2 }

or, (13)^{2} = (5)^{2} + OA^{2} [PA = 13 cm, OP = 5 cm]

or, 169 = 25 + OA^{2} or, OA^{2} = 169 – 25 or, OA^{2} = 144 or, OP = 12

∴ AB = 2 x OA [OA = OB]

or, AB = 2 x 12 cm or, AB = 24 cm.

Hence the distance between the centres of the circles is 24 cm.

**Class 10 Maths Wbbse Solutions**

**Example 2. The lengths of two chords AB and CD of a circle with its centre O are equal. If ∠AOB = 60° and CD = 5 cm, then calculate the length of the radius of the circle.**

Solution:

**Given:**

The lengths of two chords AB and CD of a circle with its centre O are equal. If ∠AOB = 60° and CD = 5 cm

In ΔAOB, OA = OB [radii of same circle] ∴∠OAB = ∠OBA

Now, ∠AOB + ∠OAB + ∠OBA = 180°

or, 60° + ∠OAB + ∠OAB = 180° [∠OBA = ∠OAB]

or, ∠OAB + ∠OAB = 180° – 60°

or, 2 ∠OAB = 120° or ∠OAB = 60°

In ΔOAB, ∠OAB = ∠OBA = ∠AOB = 60°

∴ ΔAOB is. equilateral, ∴ OA = OB = AB

But AB = CD – 5 cm, ∴ OA = OB = 5 cm

∴ Radius of the circle = 5 cm

**Example 3. R is any point in a circle with its centre O. If the length of the radius is 10 cm and OR = 6cm, then determine the least of the chord passing through the point R.**

Solution:

**Given:**

R is any point in a circle with its centre O. If the length of the radius is 10 cm and OR = 6cm

Let PQ is a chord passing through R of the circle with centre at O.

Let us join O, R.

According to the question, OP = 10 cm and OR 6cm

Now, length of PQ will be least if OR is a perpendicular to PQ.

∴ ΔOPR is a right angled triangle.

∴ OP^{2} = PR^{2} + OR^{2} or, (10)^{2} = PR^{2} + (6)^{2}

or, 100 = PR^{2} + 36 or, PR^{2} =100 – 36 or, PR^{2} = 64

∴ PR = √64 =8

Again, OR ⊥ PQ, ∴ R is the mid-point of PQ.

∴ PQ = 2 x PR = 2 x 8 cm = 16 cm.

Hence the required least length = 16 cm.

**Example 4. The two circles with their centres at P and Q intersect each other, at the points A and B. Through point A, a straight line parallel to PQ intersects the two circles at the points C and D respectively. If PQ = 7 cm, then determine the length of CD.**

Solution:

**Given:**

The two circles with their centres at P and Q intersect each other, at the points A and B. Through point A, a straight line parallel to PQ intersects the two circles at the points C and D respectively. If PQ = 7 cm

Let PM ⊥ CD and QN _⊥ CD.

∴ PQ ∥ CD, ∴ PQ = MN.

Now, PQ = 7 cm, ∴ MN = 7 cm.

Also, MN = AM + AN

or, MN = \(\frac{1}{2}\) AC + \(\frac{1}{2}\) AD [PM ⊥ AC ⇒ AM = \(\frac{1}{2}\) AC and

QN ⊥ AD ⇒ AN = \(\frac{1}{2}\) AD.] or, MN = \(\frac{1}{2}\) (AC + AD)

or, 7 = \(\frac{1}{2}\) x CD [AC + AD = CD] or, CD = 7×2 = 14

∴ the length of the chord CD = 14 cm.

**Example 5. AB is a diameter and ∠ACB is a Semicircular angle of a circle of radius 4 cm in length. If BC = 2√7 cm, find the length of AC. **

Solution:

**Given:**

AB is a diameter and ∠ACB is a Semicircular angle of a circle of radius 4 cm in length. If BC = 2√7 cm

∠ACB = 90°

In ΔACB, AC^{2} + BC^{2} = AB^{2 }⇒ AC^{2}=AB^{2}-BC^{2}

= {(2×4)^{2}-(2√7)^{2}} cm^{2}

⇒ AC = \(\sqrt{64-28}\) cm = √36 cm = 6 cm

**Example 6. If the ratio of three consecutive angles of a cyclic quadrilateral is 1 : 2 : 3 then determine the first and third angles. **

Solution:

**Given:**

If the ratio of three consecutive angles of a cyclic quadrilateral is 1 : 2 : 3

First angle = x°,

Second angle = 2x° and third angle = 3x°. [x is common multiple and x > 0]

The opposite angles a circle quadrilateral are supplementary.

∴ x° + 3x° = 180°

⇒ 4x° = 180°

⇒ x° = \(\frac{180^{\circ}}{4^{\circ}}\)

∴ First angle is 45° [First angle = x° as per question]

The measurement of third angle is 3x° = 3 x 45^{0} = 135° [Third angle = 3x° as per question]

∴ First angle is 45° and third angle is 135°.

**Example 7. Side AB of a cyclic quadrilateral ABCD is produced to the points if ∠XBC = 82° and ∠ADB = 47°, then the find the value of ∠BAC.**

Solution:

**Given:**

Side AB of a cyclic quadrilateral ABCD is produced to the points if ∠XBC = 82° and ∠ADB = 47°

In cyclic quadrilateral ABCD.

Exterior ZXBC = interior opposite ∠ABC = 82° = ∠ADC – ∠BDC = ∠ADC – ∠ADB

= 82° – 47° = 35°

∠BAC = ∠BDC [angles in the same segment] = 35°

∴ The value of ∠BAC = 35°

**Example 8. If PQRS is a cyclic parallelogram, then find the value of ∠QPR.**

Solution:

In cyclic parallelogram PQRS

∠P + ∠R=180° .

Again,∠P + ∠R [as PQRS is parallelogram]

∠P +∠P = 180° or, 2 ∠A = 180°

or, ∠A = 90° [ cyclic parallelogram is a rectangular]The value of ∠QPR = 90°

## Solid Geometry Chapter 1 Theorems Related To Circle Long Answer Type Questions

**Example 1. If the length of a chord of a circle is 48 cm and the distance of this chord from the centre is 7 cm, then find the length of the chord, the distance of which from the centre is 15 cm.**

Solution:

**Given:**

If the length of a chord of a circle is 48 cm and the distance of this chord from the centre is 7 cm

Let AB and PQ are two chords of the circle with the centre at O.

OC and OD are two perpendiculars drawn from O to AB and PQ respectively.

As per the question, OC = 7 cm

AB = 48 cm, OD = 15 cm, PQ = ?

Now, AB = 48 cm, ∴ BC = \(\frac{1}{2}\) x 48 cm = 24 cm

Then, from the right-angled triangle OBC we get,

OB^{2} = BC^{2} + OC^{2} or, OB^{2} = (24)^{2} + (7)^{2}

or, OB^{2} = 576 + 49 or, OB^{2} = 625 or, OB = √625 = 25

OQ = 25 cm [OB = OQ (radii of same circle)]

Again, in right-angled triangle OQD, OQ^{2} = OD^{2} + DQ^{2 }

or, (25)^{2} = (15)^{2} + DQ^{2} or, 625 = 225 + DQ^{2}

or, DQ^{2} = 625 – 225 or, DQ^{2} = 400 or, DQ = √400 = 20

∴ PQ = 2DQ = 2 x 20 cm = 40 cm.

Hence length of the chord PQ = 40 cm.

**Example 2. The two circles with centres X and Y intersect each other at points A and B, A is joined with the mid-point S of XY and the perpendicular on SA through the point A is drawn which intersects the two circles at the points P and Q. Prove that PA = AQ.**

Solution:

**Given:**

The two circles with centres X and Y intersect each other at points A and B, A is joined with the mid-point S of XY and the perpendicular on SA through the point A is drawn which intersects the two circles at the points P and Q.

Let us draw two perpendiculars XM and YN on PA and AQ respectively.

Let us join X, A and Y, A

XM ⊥ PA, ∴AM = \(\frac{1}{2}\) PA…..(1)

Again, YN ⊥ AQ ∴ AN = \(\frac{1}{2}\) AQ…..(2)

Now, S is the mid-point XY,

∴ AS ⊥ XY [the line segment joining two centres of two intersecting circles bisect their common chord in right angles.]

∴ in right-angled triangles AXS and AYS, we get, XS = YS [S is the mid-point of XY]

∴ ∠ASX = ∠ASY [each is right angle] and AS is common to both.

∴ ΔAXS ≅ ΔAYS, XA = YA [similar sides of two congruent triangles]’ Now, XY and PQ are both perpendiculars to AS.

∴ XY ∥ PQ.

Again, XM and YN are both perpendicular to PQ,

∴ XM = YN

Now, in ΔAXM and ΔAYN, XA = YA, SX = SY and SA is common to both.

∴ ΔAXM ≅ΔAYN [by the S-S-S condition of congruency]

∴ AM = AN [similar sides of congruent triangles]

or, \(\frac{1}{2}\) PA = \(\frac{1}{2}\) AQ [from (1) and (2)]

or, PA = AQ, ∴ PA = AQ (Proved)

**Maths Solutions Class 10 Wbbse**

**Example 3. If the angle-bisector of two intersecting chords of a circle passes through its centre, then prove that the two chords are equal.**

Solution:

**Given:**

If the angle-bisector of two intersecting chords of a circle passes through its centre,

Let the two chords AB and AC of the circle with centre at O intersect each other at A.

The bisector AO of the internal angle ∠BAC passes through the centre O.

We have to prove that AB = AC.

** Construction**: Let us join O, B and O, C.

** Proof**: In ΔAOB and ΔAOC,

∠OBA = ∠OAB [OB = OA = radii of same circle]

= ∠OAC [∠OAB = ∠OAC, given ]

= ∠OCA [OC = OA = radii of same circle ] i..e., ∠OBA = ∠OCA and ∠OAB = ∠OAC and OA is common to both.

By the condition of A-A-S congruency of congruent triangles, ΔAOB ≅ ΔAOC

∴AB = AC [similar sides of congruent triangles.]

∴ the chords AB and AC are equal. (Proved) .

**Example 4. The two parallel chords AB and CD with the lengths of 10 cm and 24cm in a cirlce are situated on the opposite sides of the center. If the distance between two chords AB and CD is 17cm, then calculate the length of the radius of the circle.**

Solution:

**Given:**

The two parallel chords AB and CD with the lengths of 10 cm and 24cm in a cirlce are situated on the opposite sides of the center. If the distance between two chords AB and CD is 17cm,

Let the length of the chord AB is 10 cm of the circle with centre at O and that of CD is 24 cm.

PQ is a straight line through O, which is perpendicular to both AB and CD.

∴ OP ⊥ AB ⇒ P is the mid-point of AB.

∴ AP = \(\frac{1}{2}\) AB = \(\frac{1}{2}\) x 10cm = 5cm

Again, OQ ⊥ CD ⇒Q is the mid-point of CD.

∴ DQ = \(\frac{1}{2}\) CD = \(\frac{1}{2}\) x 24 cm = 12 cm .

As per question, PQ = 17 cm, ∴ if OP = x cm, then OQ = (17 – x) cm.

Now, from the right-angled triangle OAP, we get, OA^{2} = AP^{2} + OP^{2}

or, OA^{2} = (5)^{2} + (x)^{2} or, OA^{2} = 25 + x^{2} …….(1)

Again, from OQD right-angled triangle we get, OD^{2} = DQ^{2} + OQ^{2}

or, OD^{2} = (12)^{2} + (17 – x)^{2} or, OD^{2} = 144 + (17 – x)^{2}

or, OA^{2} = 144 + (17 – x)^{2} (2) [OD = OA = radii of same circle]

Then from (1) and (2) we get, 144 + (17 – x)^{2} = 25 + x^{2}

or, (17 – x)^{2} – x^{2} = 25 – 144 or, (17 – x + x) (17 – x – x) = – 119

or, 17 (17 – 2x) = – 119 or, 17 – 2x = – 7 or, 2x = 24 or, x = 12

from (1) we get, OA^{2} = 25 + (12)^{2} = 25 + 144 = 169

∴ OA = √l69 = 13

Hence the required radius of the circle = 13 cm.

**Maths Solutions Class 10 Wbbse**

**Example 5. The two chords AB and AC of a circle are equal. Prove that the bisector of ∠BAC passes through the centre.**

Solution:

**Given:**

The two chords AB and AC of a circle are equal.

Let AB and AC are two equal chords of the circle with centre at O.

Let us join A, O and let it be produced to D.

We have to prove that AD passes through O, i.e it is sufficient to prove that AD is the bisector of ∠BAC.

** Construction**: Let us join O, B and O, C. ^

** Proof**: In ΔAOB and ΔAOC, OB = OC (radii of same circle) AB = AC (given) and OA is common to both.

∴ ΔAOB ≅ ΔAOC [by the condition of S-S-S congruence of triangles]

∴ ∠OAB = ∠OAC [similar angles of congruent triangles]

∴ AO or AD is the bisector of ∠BAC

Hence the bisector of ∠BAC passes through the centre. (Proved)

**Example 6. Write with proof which of the chords passing through any point in a circle will be the least.**

Solution:

Infinite number of chords can be drawn through a point in a circle.

Among them the chord of which the line segment obtained by joining the centre and the internal point is a perpendicular bisector will be the least in length.

Because, we know that so far as the perpendicular distance of the chords from the centre increases, the length of the chord proportionately diminishes and among the chords which can be drawn through this point, the perpendicular distance from the centre of that chord will be the greatest, i.e., the perpendicular distance of that chord from the centre will be the greatest.

Hence the length of that very chord will be the least.

Mathematically, let P be any point in the circle with centre at O. Let us join O, P.

Let us draw a line segment CD such that CD is perpendicular to OP at P and which intersects the circle at C and D.

Then CD is a chord passing through P. Let AB be another chord passing through P.

We have to prove that the chord CD is smaller.

** Construction**: Let us draw OQ ⊥ AB

** Proof**: ∠OQP = 1 right angle [OQ ⊥ AB]

In ΔOPQ, ∠OQP > ∠OPQ ⇒ OP > OQ.

[the side opposite to greater angle is greater than the side opposite to smaller angle.]

i.e.., the perpendicular distance of the chord CD from the centre O is greater than that of the chord AB.

So, the chord CD is at a greater distance than that of AB from the centre.

∴ the chord CD is smaller than the chord AB.

In a similar way, it can be proved that all the chords passing through P are greater than the chord CD.

Hence CD is the least chord. (Proved)

**Example 7. The centres of two circles are P and Q; they intersect at points A and B. The straight line parallel to the line segment PQ through the point A intersects the two circles at points C and D. Prove that CD = 2PQ.**

Solution:

**Given:**

The centres of two circles are P and Q; they intersect at points A and B. The straight line parallel to the line segment PQ through the point A intersects the two circles at points C and D.

Let us draw PM ⊥ CA and QN ⊥ AD

∴ AM = \(\frac{1}{2}\) CA and AN = \(\frac{1}{2}\) AD

∴ AM + AN = \(\frac{1}{2}\) CA + \(\frac{1}{2}\) AD or, MN = \(\frac{1}{2}\) (CA + AD)

or, MN = \(\frac{1}{2}\) CD ………(1)

Again, CD | | PQ and PM ⊥ CD and QN ⊥ CD, ∴ PQ = MN

∴ PQ = \(\frac{1}{2}\) CD [from (1)] or, CD = 2PQ

∴ CD = 2PQ (proved)

**Maths Solutions Class 10 Wbbse**

**Example 8. In the circle of the adjoining with its centre at O, OP ⊥ AB; if AB = 6 cm and PC = 2 cm, ****then find the length of radius of the circle.**

Solution:

**Given:**

In the circle of the adjoining with its centre at O, OP ⊥ AB; if AB = 6 cm and PC = 2 cm,

Let us join O, A. Then the radius of the circle is OA.

Now, AB = 6 cm,

∴ AP = \(\frac{1}{2}\) AB, [p is mid-point of AB]

or, AP = \(\frac{1}{2}\) x 6 cm

∴ AP = 3cm……..(1)

Again, in the right-angled triangle AOP, OA^{2} = AP^{2} + OP^{2}

or, OA^{2} = (3)^{2} + (OC – 2)^{2} [OP = OC – 2]

or, OA^{2} = 9 + OC^{2} – 4 OC + 4

or, OA^{2} = 13 + OA^{2} – 4 OC

or, 0 = 13-4 OC or, 4 OC = 13 or, OC = \(\frac{13}{4}\) = 3.25

Hence the radius = 3.25 cm.

**Example 9. The length of the radius of a circle with its centre is 6 cm and the length of its chord MN is 10 cm. Calculate the distance of the chord MN from the centre P.**

Solution:

**Given:**

The length of the radius of a circle with its centre is 6 cm and the length of its chord MN is 10 cm.

Let PO ⊥ MN

∴ distance of chord MN from P = PO.

∴ O is the mid-point of MN.

∴ MO = \(\frac{1}{}\) MN = \(\frac{1}{2}\) x 10 cm = 5 cm.

Now, in the right-angled triangle PMO PM^{2} = PO^{2} + MO^{2}

or, (10)^{2} + PO^{2} + (6)^{2} or, 100 = PO^{2} + 36

or, PO^{2} = 64 or, PO = 8 .

∴ the required distance of MN from the correct P = 8 cm.

**Example 10. The length of the diameter of a circle with its centre at O is 34 cm. The distance of the chord AB from the point O is 8 cm. Calculate the length of the chord AB**.

Solution:

**Given:**

The length of the diameter of a circle with its centre at O is 34 cm. The distance of the chord AB from the point O is 8 cm.

Let OC ⊥ AB

∴ C is the mid-point of AB. i.e. AC = \(\frac{1}{2}\) AB or, AB =2AC

∴ OC ⊥AB, ∴distance of AB from O = OC = 8 cm

Now, diameter of the circle = 34 cm

∴ Now Radius = \(\frac{34}{2}\) cm =17 cm.

Also, in right-angled mangle OAC we get,

OA^{2} = OC^{2} + AC^{2}

or, (17)^{2} = (8)^{2} + AC^{2}, or, 289 = 64 + AC^{2}

or, AC^{2} = 289 – 64 = 225

⇒ AC = √225 = 15

∴ AB = 2AC = 2 x 15 cm = 30 cm

∴ Length of the chord AB = 30 cm