PSEB Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6

PSEB Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6

PSEB 9th Class Maths Solutions Chapter 10 Circles Ex 10.6

Question 1.
Prove that the line segment joining the centres of two intersecting circles subtends equal angles at the two points of intersection.
Answer:

Circles with centres O and P intersect each other at points A and B.
In ∆ OAP and ∆ OBR
OA = OB (Radii of circle with centre O)
PA = PB (Radii of circle with centre P)
OP = OP (Common)
∴ By SSS rule, ∆ OAP = ∆ OBP
∴ ∠OAP = ∠OBP (CPCT)
Thus, OP subtends equal angles at A and B. Hence, the line segment joining the centres of two intersecting circles subtends equal angles at the two points of intersection.

Question 2.
Two chords AB and CD of lengths 5 cm and 11 cm 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.
Answer:

Draw the perpendicular bisector of AB to intersect AB at M and draw the perpendicular bisector of CD to intersect CD at N.
Both these perpendicular bisectors pass through centre O and since AB || CD; M, O and N are collinear points.
Now, MB = 1/2AB = 5/2 = 2.5 cm,
CN = 1/2CD = 11/2 = 5.5 cm and MN = 6 cm.
Let ON = x cm s
∴ OM = MN – ON = (6 – x) cm
Suppose the radius of the circle is r cm.
∴ OB = OC = r cm
In ∆ OMB, ∠M = 90°
∴OB² = OM² + MB²
∴ r² = (6 – x)² + (2.5)²
∴ r² = 36 – 12x + x² + 6.25 ………….. (1)
In ∆ ONC, ∠N = 90°
∴ OC² = ON² + CN²
∴ r² = (x)² + (5.5)²
∴ r² = x² + 30.25 ………………. (2)
From (1) and (2),
36 – 12x + x² + 6.25 = x² + 30.25
∴ – 12x = 30.25 – 6.25 – 36
∴- 12x = – 12
∴x = 1
Now, r² = x² + 30.25
∴ r² = (1)2 + 30.25
∴ r² = 31.25
∴ r = √31.25 (Approximately 5.6)
Thus, the radius of the circle is √31.25 (approximately 5.6) cm.

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

Question 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 ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Answer:

Question 5.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Answer:

ABCD is a rhombus and its diagonals intersect at M.
∴ ∠BMC is a right angle.
A circle is drawn with diameter BC.
There are three possibilities for point M:
(1) M lies in the interior of the circle,
(2) M lies in the exterior of the circle.
(3) M lies on the circle.
According to (1), if M lies in the interior of the circle, then BM produced will intersect the circle at E. Then, ∠BEC is an angle in a semicircle and hence a right angle, i.e.,
∠MEC = 90°.
In ∆ MEC, ∠ BMC is an exterior angle.
∴ ∠ BMC > ∠ MEC, i.e., ∠ BMC > 90°. In this situation, ∠ BMC is an obtuse angle which contradicts that ∠ BMC = 90°.
Similarly, according to (2), if M lies in the exterior of the circle, then ∠BMC is an acute angle which contradicts that ∠BMC 90°. Thus, possibilities (1) and (2) cannot be true.
Hence, only possibility (3) is true, i.e., M lies on the circle.
Thus, the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.

Question 6.
ABCD is a parallelogram. The circle through A, B and C intersects CD (produced if necessary at) E. Prove that AE = AD.
Answer:

Here, the circle through A, B and C intersects CD at E.
∴ Quadrilateral ABCE is cyclic.
ABCD is a parallelogram.
∴ ∠ABC = ∠ADC
∴ ∠ABC = ∠ADE
In cyclic quadrilateral ABCE,
∠ABC + ∠AEC = 180°
∴ ∠ADE + ∠AEC = 180° ……………… (1)
Moreover, ∠AEC and ∠AED form a linear pair.
∴ ∠AED + ∠AEC = 180° ………………. (2)
From (1) and (2),
∠ADE + ∠AEC = ∠AED + ∠AEC
∴ ∠ ADE = ∠ AED
Thus, in ∆ AED, ∠ADE = ∠AED.
∴ AE = AD (Sides opposite to equal angles)
Note: If the circle intersect CD produced, l then also the result can be proved in similar way.

Question 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.
Answer:

Chords AC and BD of a circle bisect each other at point O.
Hence, the diagonals of quadrilateral ABCD bisect each other.
∴ Quadrilateral ABCD Is a parallelogram.
∴ ∠BAC = ∠ACD (Alternate angles formed by transversal AC of AB || CD)
Moreover, ∠ACD = ∠ABD (Angles in same segment)
∴ ∠BAC = ∠ABD
∴ ∠BAO = ∠ABO
∴ In A OAB, OA = OB.
But, OA = OC and OB = OD
∴ OA = OB = OC = OD
∴ OA + OC = OB + OD
∴ AC = BD
Thus, the diagonals of parallelogram ABCD are equal.
∴ ABCD is a rectangle.
∴ ∠ABC = 90°
Hence, ∠ABC is an angle in a semicircle and AC is a diameter.
Similarly, ∠BAD = 90°.
Hence, ∠BAD is an angle in a semicircle and BD is a diameter.

Question 9.
Two congruent circles intersect each Other at points A and B. Through A any line segment PAQ is drawn so that P 9 lie-on. , the two circles. Prove that BP = BQ.
Answer:

Two congruent circles with centres X and Y intersect at A and B.
Hence, AB is their common chord.
In congruent circles, equal chords subtend equal angles at the centres.
∴ ∠AXB = ∠AYB
In the circle with centre X, ∠AXB = 2∠APB and in the circle with centre Y, ∠AYB = 2∠AQB.
∴ 2∠ APB = 2∠ AQB
∴ ∠APB = ∠AQB
∴ ∠QPB = ∠PQB
Thus, in ∆ BPQ, ∠QPB = ∠PQB
∴ QB = PB (Sides opposite to equal angles)
Hence, BP = BQ.

Question 10.
In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
Answer:

In ∆ ABC, the bisector of ∠A intersects the circumcircle of ∆ ABC at D.
∴∠BAD = ∠CAD
Aso, ∠BAD = ∠BCD and ∠CAD = ∠CBD (Angles in same segment)
∴ ∠BCD = ∠CBD
Thus, in ∆ BCD, ∠BCD = ∠CBD
∴BD = CD (Sides opposite to equal angles)
Thus, point D is equidistant from B and C.
Hence, D is a point on the perpendicular bisector of BC.
Thus, the bisector of ∠ A and the perpendicular bisector of side BC intersect at D and D is a point on the circumcircle of ∆ ABC.
Thus, in ∆ ABC, if the angle bisector of ∠A and the perpendicular bisector of side BC intersect, they intersect on the circumcircle of ∆ ABC.
Note: In ∆ ABC, if AB = AC, then the bisector of ∠A and the perpendicular bisector of side BC will coincide , and would not intersect in a single point.

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