Circles Exercise Ex. 8A (By:- Ajay Raj)
Solution 1
PA is the tangent to the circle with center O and radius AO = 8 cm. The point P is at a distance of 17 cm from O.
In 
PAO, 
A = 90
Solution 2
PA is the tangent to the circle with centre O and radius, such that PO = 25 cm, PA = 24 cm
In 
PAO, 
A = 90,
By Pythagoras theorem:
Hence, the radius of the circle is 7 cm.
Solution 3
Solution 4
Solution 5
Given AP is a tangent at A and OA is radius through A and PA and PB are the tangent segments to circle with centre O.
Therefore, OA is perpendicular to AP, similarly, OB is perpendicular to BP.
OAP = 90
And OBP = 90
So, OAP = OBP = 90
OBP + OAP = (90 + 90) = 180
Thus, the sum of opposite angles of quad. AOBP is 180
AOBP is a cyclic quadrilateral
Solution 6
Solution 7
Given: From an external point P, tangent PA and PB are drawn to a circle with centre O. CD is the tangent to the circle at a point E and PA = 14cm.
Since the tangents from an external point are equal, we have
PA = PB,
Also, CA = CE and DB = DE
Perimeter of 
PCD = PC + CD + PD
=(PA - CA) + (CE + DE) +(PB - DB)
= (PA - CE) + (CE + DE) + (PB - DE)
= (PA + PB) = 2PA = (2 14) cm
= 28 cm
Hence, Perimeter of 
PCD = 28 cm
Solution 8
A circle is inscribed in a triangle ABC touching AB, BC and CA at P, Q and R respectively.
Also, AB = 10 cm, AR = 7cm, CR = 5cm
AR, AP are the tangents to the circle
AP = AR = 7cm
AB = 10 cm
BP = AB - AP = (10 - 7)= 3 cm
Also, BP and BQ are tangents to the circle
BP = BQ = 3 cm
Further, CQ and CR are tangents to the circle
CQ = CR = 5cm
BC = BQ + CQ = (3 + 5) cm = 8 cm
Hence, BC = 8 cm
Solution 9
Let the circle touches the sides AB, BC, CD and DA at P, Q, R, S respectively
We know that the length of tangents drawn from an exterior point to a circle are equal
AP = AS ----(1) {tangents from A}
BP = BQ ---(2) {tangents from B}
CR = CQ ---(3) {tangents from C}
DR = DS----(4) {tangents from D}
Adding (1), (2) and (3) we get
AP + BP + CR + DR = AS + BQ + CQ + DS
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
AD = (AB + CD) - BC = {(6 + 4) - 7} cm = 3 cm
Hence, AD = 3 cm
Solution 10
Solution 11
Given O is the centre of two concentric circles of radii 4 cm and 6 cm respectively. PA and PB are tangents to the outer and inner circle respectively. PA = 10cm. Join OA, OB and OP.
Then, OB = 4 cm, OA= 6 cm and PA = 10 cm
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
