Note: In almost all the problems below the implementation of the following theorem is important. Theorem 3: The perpendicular from the center of a circle to a chord bisects the chord.

Question 1: The radius of the circle is 13 \ cm and the length of one of its chords is 10 \ cm . Find the distance of the chord from the center.2018-11-29_8-05-06

Answer:

Refer to the adjoining diagram.

OL = \sqrt{OA^2 - AL^2} = \sqrt{13^2 - 5^2} = \sqrt{144} = 12 cm

\\

Question 2: The radius of the circle is 8 \ cm and the length of one of its chords is 12 \ cm . Find the distance of the chord from the center.2018-11-29_8-05-17

Answer:

Refer to the adjoining diagram.

OL = \sqrt{8^2 - 6^2} = \sqrt{64 - 36} = \sqrt{28} = 2\sqrt{7} cm

\\

Question 3: Find the length of a chord which is at a distance of 5 \ cm from the center of the circle of radius 13 \ cm .2018-11-29_19-58-20

Answer:

Refer to the adjoining diagram.

AL = \sqrt{13^2 - 5^2} = \sqrt{144}  = 12 cm

Therefore AB = 2 \times 12 = 24 cm

\\

Question 4: Find the length of a chord which is at a distance of 5 \ cm from the center of the circle of radius 10 \ cm .2018-11-29_8-04-40

Answer:

Refer to the adjoining diagram.

AL = \sqrt{10^2 - 5^2} = \sqrt{75} = 5\sqrt{3} cm

Therefore AB = 2 \times 5\sqrt{3} = 10\sqrt{3} cm

\\

Question 5: Find the length of a chord which is at a distance of 4 \ cm from the center of the circle of radius 6 \ cm .2018-11-29_8-04-53

Answer:

Refer to the adjoining diagram.

AL = \sqrt{6^2 - 4^2} = \sqrt{36-16} = \sqrt{20} = 2\sqrt{5} cm

Therefore AB = 2 \times 2\sqrt{5} = 4\sqrt{5} cm

\\

Question 6: Two chords AB and CD of length 5 \ cm and 11 \ cm respectively of a circle are parallel. If the distance between AB and CD is 3 \ cm , find the radius of the circle.2018-11-29_8-04-07

Answer:

Refer to the adjoining diagram.

OM = \sqrt{r^2 - 5.5^2}

OL = \sqrt{r^2 - 2.5^2}

OL - OM = 3

\sqrt{r^2 - 2.5^2} - \sqrt{r^2 - 5.5^2} = 3

\Rightarrow \sqrt{r^2 - 2.5^2}  = 3 + \sqrt{r^2 - 5.5^2}

Squaring both sides we get

r^2 - 2.5^2 = 9 + r^2 - 5.5^2 + 6 \sqrt{r^2 - 5.5^2}

\Rightarrow 15 = 6 \sqrt{r^2 - 5.5^2}

\Rightarrow 5 = 2 \sqrt{r^2 - 5.5^2}

Squaring both sides we get

25 = 4(r^2 - 5.5^2)

\Rightarrow r^2 = \frac{25}{4} + 5.5^2

\Rightarrow r = \sqrt{\frac{25}{4} + 5.5^2} = \sqrt{\frac{146}{4}} = \frac{\sqrt{146}}{2}

\\

Question 7: An equilateral triangle of side 9 \ cm is inscribed in a circle. Find the radius of the circle.2018-11-30_8-24-19

Answer:

Refer to the adjoining diagram.

AL = \sqrt{9^2 - 4.5^2} = \sqrt{81-20.05} = \sqrt{60.75}

OL = \sqrt{60.75} - r

OB^2 = OL^2 + BL^2

r^2 = (\sqrt{60.75} - r)^2 + 4.5^2

\Rightarrow r^2 = 60.75 + r^2 - 2\sqrt{60.75} - r + 4.5^2

\Rightarrow 2 \sqrt{60.75} \ r = 60.75 + 4.5^2

\Rightarrow 2 \sqrt{60.75} \ r = 81

\Rightarrow r = \frac{81}{2\sqrt{60.75}} = 5.196 cm

\\

Question 8: AB is a diameter of the circle. M is a point in AB such that AM = 18 \ cm and MB = 8 \ cm . Find the length of the shortest chord through M .2018-11-29_8-03-25

Answer:

Refer to the adjoining diagram.

r = \frac{18+8}{2} = 13 cm

AM = 18 cm

AO = 13 cm

Therefore OM = 5

Therefore MC = \sqrt{13^2- 5^2} = 12 cm

Hence CD = 2 \times 12 = 24 cm

\\

Question 9: The length of the common chord of two intersecting circles is 30 \ cm . If the radii of the two circles are 25 \ cm and 17 \ cm , find the distance between their centers. 2018-11-29_8-03-09

Answer:

Refer to the adjoining diagram.

QR = r_1 \ \ \ \ \      OA = 17  \ \ \ \ \    OQ = 17-r_1  \ \ \ \ \    AQ = 15

Therefore 17^2 = (17-r_1)^2 + 15^2

289 - 225 = (17-r_1)^2

\Rightarrow 64 = (17-r_1)^2

\Rightarrow 8 = 17 - r_1

\Rightarrow r_1 = 9 cm

Similarly, PQ = r_2

O'B = 25  \ \ \ \ \  QB = 15

Therefore QO' = 25 - r_2

\Rightarrow 25^2 = (25-r_2)^2 + 15^2

\Rightarrow 625-225 = (25-r_2)^2

\Rightarrow 20 = 25 - r_2

\Rightarrow r_2 = 5 cm

Hence OO' = OR + PO' - r_1 -r_2

\Rightarrow OO'= 17 + 25 - 9 - 5

\Rightarrow OO'= 28 cm

\\

Question 10: A rectangle with a side of length 4 \ cm is inscribed in a circle of diameter 5 \ cm . Find the area of the rectangle.2018-11-29_8-00-38

Answer:

Refer to the adjoining diagram.

OL^2 = 2.5^2 - 2^2 = 6.25 - 4 = 2.25

\Rightarrow OL = 1.5 cm

AD = 2 \times 1.5 = 3 cm

Hence the area of rectangle = 4 \times 3 = 12 \ cm^2

\\

Question 11: The center of a circle of radius 13 \ cm units is the point (3,6) . P(7,9) is a point inside the circle. APB is a chord of the circle such that AP = PB . Calculate the length of AB .2018-11-29_8-00-23

Answer:

Refer to the adjoining diagram.

OP = \sqrt{(9-6)^2+(7-3)^2} = \sqrt{9+16} = \sqrt{25} = 5 cm

Therefore AP = \sqrt{13^2 - 5^2} = 12 cm

Hence AB = 2 \times 12 = 24 cm

\\

Question 12: PQ and RS are two parallel chords of a circle whose center is O and radius is 10 \ cm . If PQ is 16 \ cm and RS is 12 \ cm , find the distance between PQ and RS , if they lie i) on the same side of center O ii) on the opposite side of center O 2018-11-29_8-00-01

Answer:

Refer to the adjoining diagram.

i) OL = \sqrt{10^2 - 8^2} = \sqrt{36} = 6 cm

OM = \sqrt{10^2 - 6^2} = \sqrt{64} = 8 cm

Therefore LM = 8 - 6 = 2 cm

ii) OL = \sqrt{10^2 - 8^2} = \sqrt{36} = 6 cm

OM = \sqrt{10^2 - 6^2} = \sqrt{64} = 8 cm

Therefore LM = 8 + 6 = 14 cm

\\

Question 13: AB and CD are two parallel chords of a circle such that AB = 10 \ cm and CD = 24 \ cm . If the chords are on the opposite sides of the center and the distance between then is 17 \ cm , find the radius of the circle.

Answer:

Refer to the adjoining diagram.2018-11-29_7-59-31

OL^2 = r^2 - 5^2

\Rightarrow OL = \sqrt{r^2 -5^2}

Similarly, OM = \sqrt{r^2 - 12^2}

OL + OM = 17

Therefore \sqrt{r^2 -5^2} + \sqrt{r^2 - 12^2} = 17

\Rightarrow \sqrt{r^2 -5^2} =  17 - \sqrt{r^2 - 12^2}

Squaring both sides

r^2 - 5^2 = 289 + r^2 - 12^2 -34\sqrt{r^2 - 12^2}

\Rightarrow 34\sqrt{r^2 - 12^2} = 289 - 144 + 25

\Rightarrow 34\sqrt{r^2 - 12^2} = 170

\Rightarrow \sqrt{r^2 - 12^2} = 5

Squaring both sides

r^2 - 12^2 = 25

\Rightarrow r^2 = 144 + 25

\Rightarrow r = 13 cm

\\

Question 14: AB and CD are two chords of a circle such that AB = 6 \ cm and CD = 12 \ cm . AB \parallel CD . If the distance between AB and CD is 3 \ cm , find the radius of the circle.

Answer:2018-11-29_7-59-44

Refer to the adjoining diagram.

OM = \sqrt{r^2 - 6^2}

OL = \sqrt{r^2 - 3^2}

OL - OM = 3

\sqrt{r^2 - 3^2} - \sqrt{r^2 - 6^2} = 3

\Rightarrow \sqrt{r^2 - 3^2}  = 3 +  \sqrt{r^2 - 6^2}

Squaring both sides

r^2 - 3^2 = 9 + r^2 - 6^2 + 6\sqrt{r^2 - 6^2}

\Rightarrow 18 = 6 \sqrt{r^2 - 6^2}

\Rightarrow 3 = \sqrt{r^2 - 6^2}

Squaring both sides

9 = r^2 - 6^2

\Rightarrow r^2 = 45

\Rightarrow r = \sqrt{45} cm

\\

Question 15: ABC is an isosceles triangle inscribed in a circle. If AB = AC = 15 \ cm and BC = 18 \ cm , find the radius of the circle.

Answer:2018-11-29_7-58-51

Refer to the adjoining diagram.

OM = \sqrt{r^2 - 9^2}

AM = \sqrt{15^2 - 9^2} = \sqrt{225 - 81} = 12 cm

r + \sqrt{r^2 - 9^2} = 12

\Rightarrow  \sqrt{r^2 - 9^2} = 12 - r

Squaring both sides

r^2 - 9^2 = 144 + r^2 - 24 r

\Rightarrow 24r = 225

\Rightarrow r = \frac{75}{8} = 9 \frac{3}{8} cm

\\

Question 16: In a circle of radius 5 \ cm , AB and AC are two chords such that AB = AC = 6 \ cm . Find the length of the chord BC .

Answer:2018-11-29_7-59-03

Refer to the adjoining diagram.

OM = \sqrt{5^2 - x^2}

AM = 5 + OM = 5 + \sqrt{5^2 - x^2}

Therefore (5 + \sqrt{5^2 - x^2})^2 + x^2 = 6^2

\Rightarrow 25 + 5^2 - x^2 + 10 \sqrt{5^2 - x^2} = 6^2

\Rightarrow 50 + 10 \sqrt{5^2 - x^2} = 36

\Rightarrow 10 \sqrt{5^2 - x^2} = -14

Squaring both sides

100 (5^2 - x^2) = 196

\Rightarrow x^2 = 25 - 1.96 = \sqrt{23.04} = 4.79 cm

\\

Question 17: Two concentric circles with center O have A, B, C, D as the points of intersection with the line l as shown in the diagram. If AD = 12 \ cm and BC = 8 \ cm , find the lengths of AB, CD, AC and BD

Answer:2018-11-29_7-58-29

Refer to the adjoining diagram.

AD = 12 cm    BC = 8 cm

AB = 2 cm      CD = 2 cm

AC = 2 + 8 = 10 cm

BD = 8 + 2 = 10 cm

\\

Question 18: Two circles of radii 10 \ cm and 8 \ cm intersect and the length of the common chord is 12 \ cm . Find the distance between their centers.

Answer:2018-11-29_7-57-49

Refer to the adjoining diagram.

(8-r_1)^2 + 6^2 = 8^2

\Rightarrow (8-r_1)^2 = 64 - 36 = 28

\Rightarrow 8 - r_1 = 2\sqrt{7}

\Rightarrow r_1 = 8 - 2\sqrt{7} = 2.71

Similarly, (10 - r_2)^2 + 6^2 = 10^2

\Rightarrow (10 - r_2)^2 = 100 - 36 = 64

\Rightarrow 10 - r_2 = 8

\Rightarrow r_2 = 2

Therefore OO' = (8-r_1) + (10 - r_2)

\Rightarrow OO'= 8 - 2.71 + 10 -2

\Rightarrow OO'= 16 - 2.71 = 13.29 cm

\\

Question 19: In the figure, two circles with center A and B and of radii 5 \ cm and 3 \ cm touch each other internally. If the perpendicular bisector of segment AB meets the bigger circle in P and Q , find the length of PQ .2018-11-29_7-57-28

Answer:

Refer to the adjoining diagram.

PB = \sqrt{5^2 - 1^2} = \sqrt{24} = 2\sqrt{6}

Therefore PQ = 2 \times 2\sqrt{6} = 4\sqrt{6} cm

\\

 

Advertisements