Laravel

Chapters For Class X- CBSE


CIRCLES
Example: Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.

C1 C2 O A B P
Solution :


We are given two concentric circles $C_1$ and $C_2$ with centre O and a chord AB of the larger circle $C_1$ which touches the smaller circle $C_2$ at the point P (see Fig. 10.8). We need to prove that AP = BP

Let us join OP. Then, AB is a tangent to $C_2$ at P and OP is its radius. Therefore, by Theorem 10.1, OP ⊥ AB

Now AB is a chord of the circle $C_1$ and OP ⊥AB. Therefore, OP is the bisector of the chord AB, as the perpendicular from the centre bisects the chord, i.e., AP = BP

Example: Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠ PTQ = 2 ∠ OPQ.

O T Q P θ 90 - θ/2


Solution : We are given a circle with centre O, an external point T and two tangents TP and TQ to the circle, where P, Q are the points of contact (see Fig. 10.9). We need to prove that 
			∠ PTQ =  2 ∠ OPQ
	Let     ∠ PTQ =  θ
	Now, by Theorem 10.2, TP = TQ. So, TPQ is an isosceles triangle.
	Therefore,   ∠ TPQ = ∠ TQP =$1/2(180° − θ)$ =$90° -1/2{θ}$
	Also, by Theorem 10.1,             ∠ OPT =  90° 
	
	So,            ∠ OPQ = ∠ OPT – ∠ TPQ =$90°-(90° -1/2{θ})$
	                     =$1/2{θ}$ = $1/2$  ∠PTQ
 This gives        ∠ PTQ =  2 ∠ OPQ


Example:Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠ PTO =  ∠ OPQ. 

























O
T
Q
P
θ
(90 - θ) 







	Let     ∠ PTO =  θ
	Now, by Theorem 10.2, TP = TQ. So, PTQ is an isosceles triangle.
	Therefore,   ∠ TPQ = ∠ TQP =$1/2(180° − 2θ)$ =$90° -{θ}$
	Also, by Theorem 10.1,             ∠ OPT =  90° 
	
	So,            ∠ OPQ = ∠ OPT – ∠ TPQ =$90°-(90° -{θ})$
	                     =${θ}$ =  ∠PTO
 This gives        ∠ PTO =  ∠ OPQ



 Example: PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T (see Fig. 10.10).
            Find the length TP.
			


















O
T
Q
P
R





		

			
Solution  :  Join  OT.  Let  it  intersect  PQ  at  the point R. Then ∆ TPQ is isosceles and TO is the angle  bisector  of  ∠ PTQ.
               
			   So,  OT  ⊥ PQ and  therefore,  OT  bisects  PQ  which  gives PR = RQ = 4 cm.
			   
			    Also,    OR =     $√{OP^2   − PR^2}$=$√{5^2   − 4^2}$ = $3$ $cm$
				
				   Now,    ∠ TPR + ∠ RPO = 90° = ∠ TPR + ∠ PTR 
				   
					So,      ∠ RPO = ∠ PTR
					
			Therefore, right triangle TRP is similar to the right triangle PRO by AA similarity.
			
			So,  ${TP}/{PO}$ = ${RP}/{RO}$  i.e ${TP}/{5}$ = ${4}/{3}$ 
			
			i.e TP= $20/3$ $cm$