Q.1. Prove that the tangents drawn at the ends of a chord of circle make equal angles with the chord.
Ans. Given: A circle with centre O, PA and PB are tangents drawn at the ends A and B on chord AB.
To prove: ∠PAB = ∠PBA
Construction: Join OA and OB.
Proof: In ΔOAB we have
OA = OB ...(i) (Radii of the same circle)
∠2 = ∠1 ...(ii) (Angles opposite to equal sides of a Δ)
Also (∠2 + ∠3 = ∠1 + ∠4) ...(iii) (Both 90° as Radius ⊥ Tangent)
Subtracting (ii) from (iii), we have
∴ ∠3 = ∠4 ⇒ ∠PAB = ∠PBA
Q.2. The tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠PCA = 110°, find ∠CBA.
Ans. ∠PCA =110°
∠CBA = ?
∠ACB = 90° [Angle in a semicircle is right angle]
∠PCB = 110°- 90° = 20°
∠PCB = ∠CAB = 20°
[Alternate segment theorem]
∠ABC + ∠CAB + ∠BCA = 180°
[∵ Sum of angles of a Δ is 180°.]
⇒ ∠ABC = 180° - 110° = 70°.
Q.3. In figure, AB and CD are common tangents to two circles of unequal radii. Prove that AB = CD.
Construction: Join AD and BC
Proof: The tangents drawn from an internal point to a circle are equal in length.
If A is external point for circle having center O'
AB = AD ...(i)
If C is external point then
BC = CD ...(ii)
Now, B is external point for circle having center O
AB = BC ...(iii)
So, from (i), (ii) and (iii), we get
AB = BC = CD
So, AB = CD Hence proved.
Q.4. From an external point P, two tangents, PA and PB are drawn to a circle with centre O. At one point E on the circle tangent is drawn which intersects PA and PB at C and D, respectively. If PA = 10 cm, find the perimeter of the triangle PCD.
PA = 10 cm.
PA = PB [If P is external point] ... (i)
[∵ From an external point tangents drawn to a circle are equal in length]
If C is external point, then CA = CE
If D is external point, then
DB = DE ...(ii)
Perimeter of triangle ΔPCD
= PC + CD + PD
= PC + CE + ED + PD
= PC + CA + DB + PD
= PA + PB
= PA + PA = 2 PA
= 2 x 10 = 20 cm. [From (i)]
Q.5. If AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in figure. Prove that ∠BAT = ∠ACB.
Ans. In ΔACB,
AC is diameter, ∠B = 90°
[In a semicircle there is always a right angle]
So, ∠ACB + ∠CAB = 90° ...(i)
[∵ sum of angles of a Δ is 180°]
OA ⊥ AT [Radius and tangent are ⊥ to each other at the point of contact]
∠OAT = 90°
∠OAB + ∠BAT = 90° ...(ii)
From (i) and (ii),
∠ACB = ∠BAT