In given Fig. , two equal circles, with centres O and O’, touch each other at X. OO’ produced meets the circle with centre O’ at A. AC is tangent to the circle with centre O, at the point C. O’D is perpendicular to AC. Find the value of
In the given figure, PQ is a chord of a circle with centre O and PT is tangent. If ∠QPT = 60o, find ∠ PRQ.
PQ is the chord of the circle and PT is tangent.
We know that the tangent to a circle is perpendicular to the radius through the point of contact.
∴ ∠ OPT = 90o
Now, Given
∠ QPT = 60o
∴ ∠ OPQ = ∠OPT - ∠QPT
⇒ ∠ OPQ = 90o - 60o = 30o
In Δ OPQ,
OP= OR (Radii of the same circle)
∠ OQP = ∠ OPQ = 30o (in a triangle, equal sides have equal angles opposite to them.)
Now,
∠ OQP + ∠OPQ + ∠POQ = 180o [Angle sum property]
⇒ 30o + 30o + ∠POQ = 180o
⇒ ∠POQ = 180o - 60o = 120o
⇒ Reflex ∠POQ = 360o -120o = 240o
We know that the angle subtended by an arc of a circle at the centre is twice the angle subtended by it any point on the remaining part of the circle.
Therefore,
Reflex ∠POQ = 2 ∠PRQ
⇒ 240o = 2 ∠PRQ
⇒ ∠PRQ = 240 / 2 = 120o
Hence, the measure of angle PRQ is 120o.
In the given figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If ∠PRQ = 120o, then prove that OR = PR + RQ.
Prove that tangents drawn at the ends of a diameter of a circle are parallel to each other.
In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA = 12 cm, QC = QD = 3 cm, then find PC + PD.
Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2 ∠OPQ.
Prove that the lengths of tangents drawn from an external point to a circle are equal.
A chord PQ of a circle of radius 10 cm subtends an angle of 60° at the centre of the circle. Find the area of major and minor segments of the circle.
If the difference between the circumference and the radius of a circle is 37cm, then using , the circumference (in cm) of the circle is:
154
44
14
7