Circles : Exercise 10.6 (Optional) (Mathematics NCERT Class 9th)

Q.1     Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Sol.

Given : : Two circles with centres A and B, which intersect each other at C and D.
To prove : $\angle ACB = \angle ADB$
Construction : Join AC, AD, BD and BC.
Proof : In $\Delta s$ ACB and ADB, we have
AC = AD [Radii of the same circle]
BC = BD [Radii of the same circle]
AB = AB [Common]
Therefore by SSS criterion of congruence,
$\Delta ACB \cong \Delta ADB$
$\Rightarrow$ $\angle ACB = \angle ADB$ [C.P.C.T.]

Q.2     Two chords AB and CD of lengths 5 cm and 11 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 6 cm, find the radius of the circle.
Sol.

Let O be the centre of the given circle and let its radius be r cm. Draw $OP \bot AB$ and $OQ \bot CD$. Since $OP \bot AB$, $OQ \bot CD$ and AB || CD. Therefore, points P, O and Q are collinear. So PQ = 6 cm.

Let OP = x. Then, OQ = (6 – x) cm.
Join OA and OC. Then, OA = OC = r.
Since the perpendicular from the centre to a chord of the circle bisects the chord.
Therefore AP = PB = 2.5 cm and CQ = QD = 5.5 cm.
In right $\Delta s$ QAP and OCQ, we have
$O{A^2} = O{P^2} + A{P^2}$ and $O{C^2} = O{Q^2} + C{Q^2}$
$\Rightarrow$ ${r^2} = {x^2} + {\left( {2.5} \right)^2}$ ............... (1)
and ${r^2} = {\left( {6 - x} \right)^2} + {\left( {5.5} \right)^2}$ .............. (2)
$\Rightarrow$ ${x^2} + {\left( {2.5} \right)^2} = {\left( {6 - x} \right)^2} + {\left( {5.5} \right)^2}$
$\Rightarrow$ ${x^2} + 6.25 = 36 - 12x + {x^2} + 30.25$
$\Rightarrow$ 12x = 60 $\Rightarrow$ x = 5
Putting x = 5 in (1), we get
${r^2} = {5^2} + {\left( {2.5} \right)^2} = 25 + 6.25 = 31.25$
$\Rightarrow$ $r = \sqrt {31.25}$ = 5.6 (approx.)
Hence, the radius of the circle is 5.6 cm (approx.)

Q.3      The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre.
Sol.

Let AB and CD be two parallel chords of a circle with centre O such that AB = 6 cm and CD = 8 cm.
Let the radius of the circle be r cm.
Draw $OP \bot AB$ and $OQ \bot CD$. Since AB || CD and $OP \bot AB$, $OQ \bot CD$.
Therefore, points O, Q and P are collinear. Clearly OP = 4 cm, and P, Q are mid-points of AB and CD respectively.
Therefore $AP = PB = {1 \over 2}AB = 3$ cm
and, $CQ = QD = {1 \over 2}CD = 4$ cm
In rt. $\angle d$ $\Delta$OAP, we have
$O{A^2} = O{P^2} + A{P^2}$
$\Rightarrow$ ${r^2} = {4^2} + {3^2} = 16 + 9 = 25$
$\Rightarrow$ r = 5
In rt. $\angle d$ $\Delta$ OCQ, we have
$O{C^2} = O{Q^2} + C{Q^2}$
$\Rightarrow$ ${r^2} = O{Q^2} + {4^2}$
$\Rightarrow$ $25 = O{Q^2} + 16$
$\Rightarrow$ $O{Q^2} = 9$
$\Rightarrow$ OQ = 3
Hence, the distance of chord CD from the centre is 3 cm.

Q.4      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 $\angle ABC$ is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Sol.

Since an exterior angle of a triangle is equal to the sum of the opposite angles.
In $\Delta$BDC, we have
$\angle ADC = \angle DBC + \angle DCB$ ............ (1)
Since angle at the centre is twice the angle at a point on the remaining part of circle.
Therefore $\angle ADC = {1 \over 2}\angle AOC$
and $\angle DCB = {1 \over 2}\angle DOE$ ................. (2)
From (1) and (2), we have
${1 \over 2}\angle AOC = \angle ABC + {1 \over 2}\angle DOE$ [Since $\angle DBC = \angle ABC$]
$\Rightarrow$ $\angle ABC = {1 \over 2}\left( {\angle AOC - \angle DOE} \right)$
Hence, $\angle ABC$ is equal to half the difference of angles subtended by the chords AC and DE at the centre.

Q.5      Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.
Sol.

Given : ABCD is a rhombus. AC and BD are its two diagonals which bisect each other at right angles.
To prove : A circle drawn on AB as diameter will pass through O.
Construction : From O draw PQ || AD and EF || AB.
Proof : AB = DC $\Rightarrow$ ${1 \over 2}AB = {1 \over 2}DC$
AQ = DP
[Since Q and P are mid-points of AB and CD]
Similarly AE = OQ
$\Rightarrow$ AQ = OQ = QB
$\Rightarrow$ A circle drawn with Q as centre and radius AQ passes through A, O and B.
The circle thus obtained is the required circle.

Q.6      ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
Sol.

In order to prove that AE = AD
i.e., $\Delta$AED is an isosceles triangle it is sufficient to prove that $\angle AED = \angle ADE$. Since ABCE is a cyclic quadrilateral.
Therefore $\angle AED + \angle ABC = 180^\circ$ ....... (1)
Now, CDE is a straight line.
$\Rightarrow$ $\angle ADE + \angle ADC = 180^\circ$ ............ (2)
[Since $\angle ADC$ and $\angle ABC$ are opposite angles of a parallellogram i.e. $\angle ADC = \angle ABC$
From (1) and (2), we get
$\angle AED + \angle ABC = \angle ADE + \angle ABC$
$\Rightarrow$ $\angle AED = \angle ADE$
Therefore In $\Delta$AED, we have
$\angle AED = \angle ADE$
$\Rightarrow$ AD = AE.

Q.7      AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters, (ii) ABCD is a rectangle.
Sol.

(i) Let AB and CD be two chords of a circle with center O.
Let they bisect each other at O.
Join AC, BD, AD and BC.
In $\Delta s$ AOC and BOD, we have
OA = OB [Since O is the mid-point of AB]
$\angle AOC = \angle BOD$ [Vert. opp. $\angle s$]
and, OC = OD [Since O is the mid-point of CD]
By SAS criterion of congruence,
$\Delta AOC \cong \Delta BOD$
$\Rightarrow$ AC = BD
$\Rightarrow$ arc(AC) = arc(BD) ............. (1)
Similarly, from $\Delta s$ AOD and BOC, we have
arc(AD) = arc(BC)
From (1) and (2), we have
arc(AC) + arc(AD) = arc(BD) + arc(BC)
$\Rightarrow$ arc(CAD) = arc(CBD)
$\Rightarrow$ CD divides the circle into two parts
$\Rightarrow$ CD is a diameter.
Similarly, AB is a diameter.
(ii) Since $\Delta AOC \cong \Delta BOD$ [Proved above]
$\Rightarrow$ $\angle OAC$ i.e., $\angle BAC = \angle OBD$ i.e., $\angle ABD$
$\Rightarrow$ AC || BD
Again, $\Delta AOD \cong \Delta COB$
$\Rightarrow$ AD || CB
$\Rightarrow$ ABCD is a cyclic parallelogram.
$\Rightarrow$ $\angle DAC = \angle DBA$ .............. (3)
[Opp. $\angle s$ of a parallelogram are equal]
Also, ACBD is a cyclic quadrilateral
Therefore $\angle DAC + \angle DBA = 180^\circ$ ................ (4)
From (3) and (4), we get
$\angle DAC = \angle DBA = 90^\circ$
Hence, ABCD is a rectangle.

Q.8      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 $90^\circ - {1 \over 2}A$, $90^\circ - {1 \over 2}B$, $90^\circ - {1 \over 2}C$.
Sol.

We have $\angle D$ = $\angle EDF$
= $\angle EDA + \angle ADF$
= $\angle EBA + \angle FCA$
[Since $\angle EDA$ and $\angle EBA$ are the angles in the same segment of the circle]
Therefore $\angle EDA = \angle EBA$. Similarly, $\angle ADF$ and $\angle FCA$ are the angles in the same segment and hence, $\angle ADF = \angle FCA$
$= {1 \over 2}\angle B + {1 \over 2}\angle C$
[Since BE is the internal bisector of $\angle B$ and CF is the internal bisector $\angle C$
$\angle D = {{\angle B + \angle C} \over 2}$
Similarly, $\angle E = {{\angle C + \angle A} \over 2}$
and $\angle F = {{\angle A + \angle B} \over 2}$
$\Rightarrow$ $\Rightarrow \angle D = {{180^\circ - \angle A} \over 2}$
$\angle E = {{180^\circ - \angle B} \over 2}$
and $\angle F = {{180^\circ - \angle C} \over 2}$
[Since $\angle A + \angle B = 180^\circ$]
$\Rightarrow$ $\angle D = 90^\circ - {{\angle A} \over 2}$,
$\angle E = 90^\circ - {{\angle B} \over 2}$
and $\angle F = 90^\circ - {{\angle C} \over 2}$
$\Rightarrow$ The angles of the $\Delta DEF$ are $90^\circ - {1 \over 2}A$, $90^\circ - {1 \over 2}B$, $90^\circ - {1 \over 2}C$.

Q.9      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.
Sol.

Let O and O' be the centre of two congruent circles.
Since AB is a common chord of these circles.
Therefore arc(ACB) = arc(ADB)
$\angle BPA = \angle BQA$
$\Rightarrow$ BP = BQ.

Q.10     In any triangle ABC, if the angle bisector of $\angle A$ and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
Given : ABC is a triangle inscribed in a circle with centre at O, E is a point on the circle such that AE is the internal bisector of $\angle BAC$ and D is the mid-point of BC.

To Prove : DE is the right bisector of BC i.e. $\angle BDE = \angle CDE = 90^\circ$.
Construction : Join BE and EC.
Sol.

In $\Delta BDE$ and $\Delta CDE$, we have
BE = CE
[Since $\angle BAE = \angle CAE$, therefore arc(BE) = arc(CE) $\Rightarrow$ chord BE = chord CE]
BD = CD [Given]
DE = DE [Common]
Therefore by SSS criterion of congruence,
$\Delta BDE \cong \Delta CDE$
$\Rightarrow$ $\angle BDE = \angle CDE$ [C.P.C.T.]
Also, $\angle BDE + \angle CDE = 180^\circ$ [Linear pari]
Therefore $\angle BDE = \angle CDE = 90^\circ$
Hence, DE is the right bisector of BC.