NCERT Solutions for Exercise 10.6 Class 9 Maths Chapter 10 - Circles

Circles Class 9 Chapter 10 Exercise: 10.6

Q1 Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Answer:

Given: Circle C(P,r) and circle C(Q,r') intersect each other at A and B.

To prove : $\mathrm{\angle }$ PAQ = $\mathrm{\angle }$ PBQ

Proof : In $\mathrm{△}$ APQ and $\mathrm{△}$ BPQ,

PA = PB (radii of same circle)

PQ = PQ (Common)

QA = QB (radii of same circle)

So, $\mathrm{△}$ APQ $\cong$ $\mathrm{△}$ BPQ (By SSS)

$\mathrm{\angle }$ PAQ = $\mathrm{\angle }$ PBQ (CPCT)

Q2 Two chords AB and CD of lengths $5cm$ and $11cm$ respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is $6cm$ , find the radius of the circle.

Answer:

Given : AB = 5 cm, CD = 11 cm and AB || CD.

To find Radius (OA).

Construction: Draw $OM\perp CDandON\perp AB$

Proof :

Proof: CD is a chord of circle and $OM\perp CD$

Thus, CM = MD = 5.5 cm (perpendicular from centre bisects chord)

and AN = NB = 2.5 cm

Let OM be x.

So, ON = 6 - x (MN = 6 cm )

In $\mathrm{△}$ OCM , using Pythagoras,

$O{C}^2=C{M}^2+O{M}^2$ .............................1

and

In $\mathrm{△}$ OAN , using Pythagoras,

$O{A}^2=A{N}^2+O{N}^2$ .............................2

From 1 and 2,

$C{M}^2+O{M}^2=A{N}^2+O{N}^2$ (OC=OA =radii)

${5.5}^2+x^2={2.5}^2+(6-x{)}^2$

$\Rightarrow 30.25+x^2=6.25+36+x^2-12x$

$\Rightarrow 30.25-42.25=-12x$

$\Rightarrow -12=-12x$

$\Rightarrow x=1$

From 2, we get

$O{C}^2={5.5}^2+1^2=30.25+1=31.25$

$\Rightarrow OC=\frac{5}{2}\sqrt{5}cm$

OA = OC

Thus, the radius of the circle is $\frac{5}{2}\sqrt{5}cm$

Q3 The lengths of two parallel chords of a circle are $6cm$ and $8cm$ . If the smaller chord is at distance $4cm$ from the centre, what is the distance of the other chord from the centre?

Answer:

Given : AB = 8 cm, CD = 6 cm , OM = 4 cm and AB || CD.

To find: Length of ON

Construction: Draw $OM\perp CDandON\perp AB$

Proof :

Proof: CD is a chord of circle and $OM\perp CD$

Thus, CM = MD = 3 cm (perpendicular from centre bisects chord)

and AN = NB = 4 cm

Let MN be x.

So, ON = 4 - x (MN = 4 cm )

In $\mathrm{△}$ OCM , using Pythagoras,

$O{C}^2=C{M}^2+O{M}^2$ .............................1

and

In $\mathrm{△}$ OAN , using Pythagoras,

$O{A}^2=A{N}^2+O{N}^2$ .............................2

From 1 and 2,

$C{M}^2+O{M}^2=A{N}^2+O{N}^2$ (OC=OA =radii)

$\Rightarrow 3^2+4^2=4^2+(4-x{)}^2$

$\Rightarrow 9+16=16+16+x^2-8x$

$\Rightarrow 9=16+x^2-8x$

$\Rightarrow x^2-8x+7=0$

$\Rightarrow x^2-7x-x+7=0$

$\Rightarrow x(x-7)-1(x-7)=0$

$\Rightarrow (x-1)(x-7)=0$

$\Rightarrow x=1,7$

So, x=1 (since $x\ne 7>OM$ )

ON =4-x =4-1=3 cm

Hence, second chord is 3 cm away from centre.

Q4 Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that $\mathrm{\angle }ABC$ is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Answer:

Given : AD = CE

To prove : $\mathrm{\angle }ABC=\frac{1}{2}(\mathrm{\angle }AOC-\mathrm{\angle }DOE)$

Construction: Join AC and DE.

Proof :

Let $\mathrm{\angle }$ ADC = x , $\mathrm{\angle }$ DOE = y and $\mathrm{\angle }$ AOD = z

So, $\mathrm{\angle }$ EOC = z (each chord subtends equal angle at centre)

$\mathrm{\angle }$ AOC + $\mathrm{\angle }$ DOE + $\mathrm{\angle }$ AOD + $\mathrm{\angle }$ EOC = ${360}^{\circ }$

$\Rightarrow x+y+z+z={360}^{\circ }$

$\Rightarrow x+y+2z={360}^{\circ }$ .........................................1

In $\mathrm{△}$ OAD ,

OA = OD (Radii of the circle)

$\mathrm{\angle }$ OAD = $\mathrm{\angle }$ ODA (angles opposite to equal sides )

$\mathrm{\angle }$ OAD + $\mathrm{\angle }$ ODA + $\mathrm{\angle }$ AOD = ${180}^{\circ }$

$\Rightarrow 2\mathrm{\angle }OAD+z={180}^{\circ }$

$\Rightarrow 2\mathrm{\angle }OAD={180}^{\circ }-z$

$\Rightarrow \mathrm{\angle }OAD=\frac{{180}^{\circ }-z}{2}$

$\Rightarrow \mathrm{\angle }OAD={90}^{\circ }-\frac{z}{2}$ .............................................................2

Similarly,

$\Rightarrow \mathrm{\angle }OCE={90}^{\circ }-\frac{x}{2}$ .............................................................3

$\Rightarrow \mathrm{\angle }OED={90}^{\circ }-\frac{y}{2}$ ..............................................................4

$\mathrm{\angle }$ ODB is exterior of triangle OAD . So,

$\mathrm{\angle }$ ODB = $\mathrm{\angle }$ OAD + $\mathrm{\angle }$ ODA

$\Rightarrow \mathrm{\angle }ODB={90}^{\circ }-\frac{z}{2}+z$ (from 2)

$\Rightarrow \mathrm{\angle }ODB={90}^{\circ }+\frac{z}{2}$ .................................................................5

similarly,

$\mathrm{\angle }$ OBE is exterior of triangle OCE . So,

$\mathrm{\angle }$ OBE = $\mathrm{\angle }$ OCE + $\mathrm{\angle }$ OEC

$\Rightarrow \mathrm{\angle }OEB={90}^{\circ }-\frac{z}{2}+z$ (from 3)

$\Rightarrow \mathrm{\angle }OEB={90}^{\circ }+\frac{z}{2}$ .................................................................6

From 4,5,6 ;we get

$\mathrm{\angle }$ BDE = $\mathrm{\angle }$ BED = $\mathrm{\angle }$ OEB - $\mathrm{\angle }$ OED

$\Rightarrow \mathrm{\angle }BDE=\mathrm{\angle }BED={90}^{\circ }+\frac{z}{2}-(90-\frac{y}{2})=\frac{y+z}{2}$

$\Rightarrow \mathrm{\angle }BDE+\mathrm{\angle }BED=y+z$ ..................................................7

In $\mathrm{△}$ BDE ,

$\mathrm{\angle }$ DBE + $\mathrm{\angle }$ BDE + $\mathrm{\angle }$ BED = ${180}^{\circ }$

$\Rightarrow \mathrm{\angle }DBE+y+z={180}^{\circ }$

$\Rightarrow \mathrm{\angle }DBE={180}^{\circ }-(y+z)$

$\Rightarrow \mathrm{\angle }ABC={180}^{\circ }-(y+z)$ ...................................................8

Here, from equation 1,

$\frac{x-y}{2}=\frac{{360}^{\circ }-y-2x-y}{2}$

$\Rightarrow \frac{x-y}{2}=\frac{{360}^{\circ }-2y-2x}{2}$

$\Rightarrow \frac{x-y}{2}={180}^{\circ }-y-x$ ...................................9

From 8 and 9,we have

$\mathrm{\angle }ABC=\frac{x-y}{2}=\frac{1}{2}(\mathrm{\angle }AOC-\mathrm{\angle }DOE)$

Q5 Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.

Answer:

Given : ABCD is a rhombus.

To prove: the circle drawn with AB as diameter passes through the point O.

Proof :

ABCD is rhombus.

Thus, $\mathrm{\angle }AOC={90}^{\circ }$ (diagonals of a rhombus bisect each other at ${90}^{\circ }$ )

So, a circle drawn AB as diameter will pass through point O.

Thus, the circle is drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.

Q6 ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that $AE=AD$ .

Answer:

Given: ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E.

To prove : AE = AD

Proof :

$\mathrm{\angle }$ ADC = $\mathrm{\angle }$ 3 , $\mathrm{\angle }$ ABC = $\mathrm{\angle }$ 4, $\mathrm{\angle }$ ADE = $\mathrm{\angle }$ 1 and $\mathrm{\angle }$ AED = $\mathrm{\angle }$ 2

$\mathrm{\angle }3+\mathrm{\angle }1={180}^{\circ }$ .................1(linear pair)

$\mathrm{\angle }2+\mathrm{\angle }4={180}^{\circ }$ ....................2(sum of opposite angles of cyclic quadrilateral)

$\mathrm{\angle }$ 3 = $\mathrm{\angle }$ 4 (oppsoite angles of parallelogram )

From 1 and 2,

$\mathrm{\angle }$ 3+ $\mathrm{\angle }$ 1 = $\mathrm{\angle }$ 2 + $\mathrm{\angle }$ 4

From 3, $\mathrm{\angle }$ 1 = $\mathrm{\angle }$ 2

From 4, $\mathrm{△}$ AQB, $\mathrm{\angle }$ 1 = $\mathrm{\angle }$ 2

Therefore, AE = AD (In an isosceles triangle ,angles oppsoite to equal sides are equal)

Q7 (i) AC and BD are chords of a circle which bisect each other. Prove that AC and BD are diameters

Answer:

Given: AC and BD are chords of a circle which bisect each other.

To prove: AC and BD are diameters.

Construction : Join AB,BC,CD,DA.

Proof :

In $\mathrm{△}$ ABD and $\mathrm{△}$ CDO,

AO = OC (Given )

$\mathrm{\angle }$ AOB = $\mathrm{\angle }$ COD (Vertically opposite angles )

BO = DO (Given )

So, $\mathrm{△}$ ABD $\cong$ $\mathrm{△}$ CDO (By SAS)

$\mathrm{\angle }$ BAO = $\mathrm{\angle }$ DCO (CPCT)

$\mathrm{\angle }$ BAO and $\mathrm{\angle }$ DCO are alternate angle and are equal .

So, AB || DC ..............1

Also AD || BC ...............2

From 1 and 2,

$\mathrm{\angle }A+\mathrm{\angle }C={180}^{\circ }$ ......................3(sum of opposite angles)

$\mathrm{\angle }$ A = $\mathrm{\angle }$ C ................................4(Opposite angles of the parallelogram )

From 3 and 4,

$\mathrm{\angle }A+\mathrm{\angle }A={180}^{\circ }$

$\Rightarrow 2\mathrm{\angle }A={180}^{\circ }$

$\Rightarrow \mathrm{\angle }A={90}^{\circ }$

BD is a diameter of the circle.

Similarly, AC is a diameter.

Q7 (ii) AC and BD are chords of a circle which bisect each other. Prove that ABCD is a rectangle.

Answer:

Given: AC and BD are chords of a circle which bisect each other.

To prove: ABCD is a rectangle.

Construction : Join AB,BC,CD,DA.

Proof :

ABCD is a parallelogram. (proved in (i))

$\mathrm{\angle }A={90}^{\circ }$ (proved in (i))

A parallelogram with one angle ${90}^{\circ }$ , is a rectangle )

Thus, ABCD is rectangle.

Q8 Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are ${90}^{\circ }-\frac{1}{2}C$ , ${90}^{\circ }-\frac{1}{2}B$ and ${90}^{\circ }-\frac{1}{2}A$

Answer:

Given : Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively.

To prove : the angles of the triangle DEF are ${90}^{\circ }-\frac{1}{2}C$ , ${90}^{\circ }-\frac{1}{2}B$ and ${90}^{\circ }-\frac{1}{2}A$

Proof :

$\mathrm{\angle }$ 1 and $\mathrm{\angle }$ 3 are angles in same segment.therefore,

$\mathrm{\angle }$ 1 = $\mathrm{\angle }$ 3 ................1(angles in same segment are equal )

and $\mathrm{\angle }$ 2 = $\mathrm{\angle }$ 4 ..................2

Adding 1 and 2,we have

$\mathrm{\angle }$ 1+ $\mathrm{\angle }$ 2= $\mathrm{\angle }$ 3+ $\mathrm{\angle }$ 4

$\Rightarrow \mathrm{\angle }D=\frac{1}{2}\mathrm{\angle }B+\frac{1}{2}\mathrm{\angle }C$ ,

$\Rightarrow \mathrm{\angle }D=\frac{1}{2}(\mathrm{\angle }B+\mathrm{\angle }C)$

$\Rightarrow \mathrm{\angle }D=\frac{1}{2}({180}^{\circ }+\mathrm{\angle }C)$

and $\Rightarrow \mathrm{\angle }D=\frac{1}{2}({180}^{\circ }-\mathrm{\angle }A)$

$\Rightarrow \mathrm{\angle }D={90}^{\circ }-\frac{1}{2}\mathrm{\angle }A$

Similarly, $\Rightarrow \mathrm{\angle }E={90}^{\circ }-\frac{1}{2}\mathrm{\angle }B$ and $\mathrm{\angle }F={90}^{\circ }-\frac{1}{2}\mathrm{\angle }C$

Q9 Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that $BP=BQ$ .

Answer:

Given: Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles.

To prove : BP = BQ

Proof :

AB is a common chord in both congruent circles.

$\therefore \mathrm{\angle }APB=\mathrm{\angle }AQB$

In $\mathrm{△}BPQ,$

$\mathrm{\angle }APB=\mathrm{\angle }AQB$

$\therefore BQ=BP$ (Sides opposite to equal of the triangle are equal )

Q10 In any triangle ABC, if the angle bisector of $\mathrm{\angle }A$ and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.

Answer:

Given :In any triangle ABC, if the angle bisector of $\mathrm{\angle }A$ and perpendicular bisector of BC intersect.

To prove : D lies on perpendicular bisector BC.

Construction: Join BD and DC.

Proof :

Let $\mathrm{\angle }$ ABD = $\mathrm{\angle }$ 1 , $\mathrm{\angle }$ ADC = $\mathrm{\angle }$ 2 , $\mathrm{\angle }$ DCB = $\mathrm{\angle }$ 3 , $\mathrm{\angle }$ CBD = $\mathrm{\angle }$ 4

$\mathrm{\angle }$ 1 and $\mathrm{\angle }$ 3 lies in same segment.So,

$\mathrm{\angle }$ 1 = $\mathrm{\angle }$ 3 ..........................1(angles in same segment)

similarly, $\mathrm{\angle }$ 2 = $\mathrm{\angle }$ 4 ......................2

also, $\mathrm{\angle }$ 1= $\mathrm{\angle }$ 2 ..............3(given)

From 1,2,3 , we get

$\mathrm{\angle }$ 3 = $\mathrm{\angle }$ 4

Hence, BD = DC (angles opposite to equal sides are equal )

All points lying on perpendicular bisector BC will be equidistant from B and C.

Thus, point D also lies on perpendicular bisector BC.