Circle — Chapter Test

In the adjoining figure, a chord PQ of a circle with center O and radius 15 cm is bisected at M by a diameter AB. If OM = 9 cm, find the lengths of :

(i) PQ

(ii) AP

(iii) BP.

Answer

(i) Given,

AB bisects PQ.

∴ OM bisects PQ.

Since, the straight line drawn from the centre of a circle to bisect a chord, which is not a diameter, is perpendicular to the chord,

∴ OM ⊥ PQ.

In right △OMP,

⇒ OP² = OM² + PM² (By pythagoras theorem)

⇒ 15² = 9² + PM²

⇒ PM² = 15² - 9²

⇒ PM² = 225 - 81

⇒ PM² = 144

⇒ PM = $\sqrt{144}$ = 12 cm.

PQ = 2PM = 24 cm.

Hence, PQ = 24 cm.

(ii) From figure,

AM = AO + OM = 15 + 9 = 24 cm.

In right △APM,

⇒ AP² = AM² + PM² (By pythagoras theorem)

⇒ AP² = 24² + 12²

⇒ AP² = 576 + 144

⇒ AP² = 720

⇒ AP = $\sqrt{720}$ = $12\sqrt{5}$ cm.

Hence, AP = $12\sqrt{5}$ cm.

(iii) From figure,

MB = OB - OM = 15 - 9 = 6 cm.

In right △MPB,

⇒ BP² = PM² + MB² (By pythagoras theorem)

⇒ BP² = 12² + 6²

⇒ BP² = 144 + 36

⇒ BP² = 180

⇒ BP = $\sqrt{180}$ = $6\sqrt{5}$ cm.

Hence, BP = $6\sqrt{5}$ cm.

The radii of two concentric circles are 17 cm and 10 cm; a line PQRS cuts the larger circle at P and S and the smaller circle at Q and R. If QR = 12 cm, calculate PQ.

Answer

Draw OM ⊥ QR.

In right △OQM,

⇒ OQ² = OM² + QM²

⇒ 10² = OM² + 6²

⇒ OM² = 100 - 36

⇒ OM² = 64

⇒ OM = $\sqrt{64}$ = 8 cm.

In right △POM,

⇒ PO² = OM² + PM²

⇒ 17² = 8² + PM²

⇒ PM² = 289 - 64

⇒ PM² = 225

⇒ PM = $\sqrt{225}$ = 15 cm.

From figure,

PQ = PM - QM = 15 - 6 = 9 cm.

Hence, PQ = 9 cm.

A chord of length 48 cm is at a distance of 10 cm from the centre of a circle. If another chord of length 20 cm is drawn in the same circle, find its distance from the center of the circle.

Answer

From figure,

Since, OM ⊥ AB, it bisects it. (As perpendicular from center to chord bisects it)

∴ AM = MB = $\dfrac{48}{2}$ = 24 cm.

In right △AOM,

⇒ AO² = OM² + AM² (By pythagoras theorem)

⇒ AO² = 10² + 24²

⇒ AO² = 100 + 576

⇒ AO² = 676

⇒ AO = $\sqrt{676}$ = 26 cm.

Radius = 26 cm.

Since, ON ⊥ CD, it bisects it. (As perpendicular from center to chord bisects it)

∴ CN = ND = $\dfrac{20}{2}$ = 10 cm.

In right △CNO,

⇒ OC² = ON² + NC² (By pythagoras theorem)

⇒ 26² = ON² + 10²

⇒ ON² = 676 - 100

⇒ ON² = 576

⇒ ON = $\sqrt{576}$ = 24 cm.

Hence, the chord of length 20 cm is at a distance of 24 cm from the centre.

In the figure (i) given below, two circles with centers C, D intersect in points P, Q. If length of common chord is 6 cm and CP = 5 cm, DP = 4 cm, calculate the distance CD correct to two decimal places.

Answer

From figure,

PM = MQ = $\dfrac{6}{2}$ = 3 cm.

In right △CMP,

⇒ CP² = CM² + PM² (By pythagoras theorem)

⇒ 5² = CM² + 3²

⇒ CM² = 25 - 9

⇒ CM² = 16

⇒ CM = $\sqrt{16}$ = 4 cm.

In right △DMP,

⇒ DP² = DM² + PM² (By pythagoras theorem)

⇒ 4² = DM² + 3²

⇒ DM² = 16 - 9

⇒ DM² = 7

⇒ DM = $\sqrt{7}$ = 2.65 cm.

CD = CM + MD = 4 + 2.65 = 6.65 cm.

Hence, CD = 6.65 cm.

In the figure (ii) given below, P is a point of intersection of two circles with centers C and D. If the st. line APB is parallel to CD, prove that AB = 2CD.

Answer

From C, D draw CM, DN perpendiculars to AB.

From figure,

MCDN is a rectangle.

∴ MN = CD (Opposite sides of rectangle are equal).

Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ MP = $\dfrac{1}{2}$AP and NP = $\dfrac{1}{2}$PB

From figure,

MN = MP + PN

= $\dfrac{1}{2}$AP + $\dfrac{1}{2}$PB

= $\dfrac{1}{2}$(AP + PB)

= $\dfrac{1}{2}$AB

∴ CD = $\dfrac{1}{2}$AB [∵ MN = CD]

⇒ AB = 2CD

Hence, proved that AB = 2CD.

In the figure (i) given below, C and D are centers of two intersecting circles. The line APQB is perpendicular to the line of centers CD. Prove that

(i) AP = QB

(ii) AQ = BP.

Answer

Let M be the point of intersection of line CD and line APQB.

(i) Since, the perpendicular to a chord from the centre of the circle bisects the chord,

∴ In circle with center D,

PM = MQ .........(1)

and

In circle with center C,

AM = MB .........(2)

Subtracting equation (1) from (2),

⇒ AM - PM = MB - MQ

⇒ AP = QB

Hence, proved that AP = QB.

(ii) Let AP = QB = x.

From figure,

AQ = AB - QB = AB - x

BP = AB - AP = AB - x.

∴ AQ = BP.

Hence, proved that AQ = BP.

In the figure (ii) given below, two equal chords AB and CD of a circle with center O intersect at right angles at P. If M and N are mid-points of the chords AB and CD respectively, prove that NOMP is a square.

Answer

In NOMP,

∠P = 90° (As chords intersect at right angles)

∠M = ∠N = 90° (Straight lines from center bisecting the chord are perpendicular to it.)

∠O = 360° - (∠M + ∠N + ∠P)
= 360° - (90° + 90° + 90°)
= 90°

Since, equal chords are equidistant from center,

∴ OM = ON.

Since, all angles = 90° and all sides are equal.

Hence, NOMP is a square.

In the adjoining figure, AD is diameter of a circle. If the chord AB and AC are equidistant from its center O, prove that AD bisects ∠BAC and ∠BDC.

Answer

As chords AB and AC are equidistant from the center, so AB = AC. [∵ equal chords are equidistant from center]

Since, angle in a semicircle = 90°.

∠B = ∠C

In △ABD and △ACD,

∠B = ∠C (Both equal to 90°)

AD = AD (Common)

AB = AC (Proved above)

∴ △ABD ≅ △ACD (By R.H.S. congruence rule).

∴ ∠BAD = ∠CAD and ∠BDA = ∠CDA (By C.P.C.T.)

Hence, proved that AD bisects ∠BAC and ∠BDC.