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The largest chord of a circle is called diameter of the circle.
Any chord which crosses the centre of the circle is the largest chord of the circle, which is the diameter of the circle.
AB and CD are two parallel chords of a circle such that AB = 8 cm, and CD = 6cm. If the chords are on the opposite sides of the center and the distance between them is 7cm, then the diameter of the circle is:
Let the radius of the circle be r, and the length OP be x
∴ In ∆OPB,
OP^{2} + PB^{2} = OB^{2} = r^{2}
⇒ x^{2} + (4)^{2} = r^{2 } [∵ PB = AB/2]
⇒ x^{2} = r^{2 }– 16 ….(i)
In ∆OCQ,
OQ^{2} + CQ^{2} = r^{2}
⇒ (7 – x)^{2} + (3)^{2} = r^{2}
[∵ OQ = PQ  OP = 7  x]
⇒ (7 – x)^{2} = r^{2} – 9 ….(ii)
Subtracting (ii) from (i), we get
x^{2} – (7 – x)^{2} = – 7
⇒ (x – 7 + x) (x + 7 – x) = – 7
⇒ (2x – 7) (7) = – 7
⇒ (2x – 7) = – 1
⇒ x = 3
∴ r^{2} = x^{2} + 16 [using (i)]
= (3)^{2} + 16 = 25
⇒ r = 5cm
∴ d = 10 cm.
Two circles of radii 13cm and 15cm intersect and the length of the common chord is 24 cm, then the distance between their centers is:
In ∆OPA,
Similarly,
In ∆O′PA,
= 9cm.
∴ OO′ = OP + O′P = 5 + 9 = 14 cm.
Bisector AD of ∠BAC of ∆ABC passes through the center of the circumcircle of ∆ABC, then,
∵ AD is the bisector of ∠BAC
∴ AD is the ⊥ bisector of BC.
∵ O is the circumcentre of ∆ABC.
∴ AB = AC [∵ ∠ABD = ∠ACB]
(By using ∆ABD ≅ ∆ACD)
Reflex ∠AOC = 360° – (110° + 120°)
= 130°
∠BOC = 2 × ∠BAC [Angle subtended at center is double the angle subtended the circle]
= 2 × 30° = 60°
P is the center of the circle, and ∠XPZ = 120°, ∠XZY = 35°, then the measure of ∠YXZ is:
∠XPY = 2∠XOP
∵ ∠XOP = ∠XZY
(angles in the same segment)
∴ ∠XPY = 2∠XZY …(i)
Similarly,
∠YPZ = 2∠YXZ …(ii)
Using (i) and (ii)
∠XPZ = 2 (∠XYZ + ∠YXZ)
PQRS is a cyclic quadrilateral such that PR is the diameter of the circle. If ∠QPR = 64° and ∠SPR = 31°, then, ∠R =?
∠P = ∠QPR + ∠SPR
= 64 ° + 31°
= 95°
∵ ∠Q and ∠S are angles of the semicircle
∴ ∠Q = ∠S = 90°
∵ PQRS is a cyclic quadrilateral
∴ ∠P + ∠R = 180°
⇒ ∠R = 180°  95° = 85°
If A, B, C are three points on a circle with center O such that ∠AOB = 90° and ∠BOC = 120°, then ∠ABC =?
Reflex ∠AOC = 360° – (90° + 120°) = 150°
O is the center of the circle, with AC = 30 cm and DA = 10√5 cm, then the measure of DC is
In ∆ACD,
∠ADC = 90° [∵∠ACD is angle in semicircle]
∴ AC^{2} = DA^{2} + DC^{2}
⇒ (30)^{2 }= (10√5 )^{2} + DC^{2}
⇒ DC^{2} = 900 – 500
⇒ DC = √400 = 20 cm.
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