Fig. 9.17 Two cireles with centres O and O of radii 3 cm and 4 cm, respectively intersect at two points P and Q such that OP and O'P are tangents to the two circles. Find the length of the common chord PQ. In a right triangle ABC in which B 90?, a circle is drawn with AB as diameter intersecting the hypotenuse AC and P. Prove that the angent to the circle at P bisects BC. RPQ a cHcc such In Fig. 9.18, tangents P? 30?. A chord RS is drawn parallel to the tangent PQ. Find the RQS.
Dear Student,
Given: ΔABC is right triangle in which ∠ABC = 90°.
A circle is drawn on side AB as diameter intersecting AC in P.
PQ is the tangent to the circle when intersects BC in Q.
Construction: Join BP.
Proof:
PQ and BQ are tangents drawn from an external point Q.
⇒ PQ = BQ ....................(i) [Length of tangents drawn from an external point to the circle are equal]
⇒ ∠PBQ = ∠BPQ [In a triangle, equal sides have equal angles opposite to them]
As , it is given that,
AB is the diameter of the circle.
∴ ∠APB = 90° [Angle in a semi-circle is 90°]
∠APB + ∠BPC = 180° [Linear pair]
⇒∠BPC = 180° – ∠APB = 180° – 90° = 90°
In ΔBPC,
∠BPC + ∠PBC + ∠PCB = 180° [Angle sum property]
⇒∠PBC + ∠PCB = 180° – ∠BPC = 180° – 90° = 90° ...................(ii)
Now,
∠BPC = 90°
⇒∠BPQ + ∠CPQ = 90° .................(iii)
From (ii) and (iii), we get,
⇒∠PBC + ∠PCB = ∠BPQ + ∠CPQ
⇒ ∠PCQ = ∠CPQ [
∠BPQ = ∠PBQ]
In ΔPQC,
∠PCQ = ∠CPQ
∴ PQ = QC ...(iv)
From (i) and (iv), we get,
BQ = QC
Thus, tangent at P bisects the side BC.
Regards