Given : Two concentric circles C_{1} and C_{2 }of radii 13 cm and 8 cm respectively. AB is a diameter of the bigger circle (C_{1}) and BD is a tangent to the smaller circle (C_{2}).
∴ The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Now, in right triangle BOD, we have
OB^{2} = OD^{2} = BD^{2} [ Using Pythagoras theorem]
(13)^{2 }= (8)^{2 }+ BD^{2 }
169 = 64 + BD^{2}
BD^{2} = 169 - 64
BD^{2} = 105
Since, perpendicular drawn from the centre to the chord bisects the chord.
(∵ ∠BEA = 90°, angle in semicircle is right angle] and, ∠OBD = ∠ABE (common)
Therefore, using AAS similar condition ΔBOD ~ ΔBAE
[Proportional sides of two similar triangles]
Now, in right triangle ADE, we have
AD^{2} = AE^{2} + DE^{2 }
= 256 + 105
= 361
AD = 19 cm.
So, BP = BQ
(Tangents from external point B)
But BP = 27 cm
⇒ BQ = 27 cm
It is given that BC = 38 cm
⇒ BQ + CQ = 38
⇒ 27 + CQ = 38
⇒ CQ = 11 cm
⇒ CQ = CR (Tagents from an external point C)
But CQ = 11 cm
⇒ CR = 11 cm
It is given that : CD = 25 cm
⇒ CR + DR = 25
⇒ 11 + DR = 25
⇒ DR = 14 cm
Since, tangent to a circle is perpendicular to the radius through the point of contact.
∴ ∠ORD = ∠OSD = 90°
It is given that
∠D = 90°
Now, in quadrilateral ORDS,
∠ORD = ∠OSD = ∠RDS = ∠ROS = 90°
and OR = OS [radii of circle]
Therefore, ORDS is a square
So, OR = DR = 14 cm
Hence r = 14 cm.
Chord
C.
SecantTherefore, ∠OPO =90°
It is given that PQ =15 cm
and OQ = 17 cm
In right ΔOPQ
OQ^{2} = OP^{2} + PQ^{2}[Using Pythagoras theorem]
⇒ (17)^{2} = OP^{2} + (15)^{2}OP^{2} = (17)^{2 }– (15)^{2}= 289 – 225 = 64 ⇒ OP = 8 cm
Hence, radius of the circle = 8 cm.
It is given that ABC is a right angle triangle with AB = 6 cm and AC = 8 cm and a circle with centre O has been inscribed.
Using Pythagoras theorem, we get
BC^{2} = AC^{2} + AB^{2}= (8)^{2} + (6)^{2}= 64 + 36 = 100
⇒ BC = 10 cm
Tangents at any point of a circle is perpendicular to the radius through the point of contact
= 3r + 4r + 5r
= 12r ...(ii)
Comparing (i) and (ii), we get
24 = 12r ⇒ r = 2 cm
Method – II:
In quadrilateral APOR,
∵ ∠OPA = ∠ORA = 90°
∠PAR = 90°
⇒ ∠OPA = ∠ORA = ∠PAR = ∠POR = 90° ....(i)
and AP = AR (ii) (length of tangents drawn from an external point are equal)
Using result (i) and (ii), we get
APOR is a square
Therefore,
OR = AR = r [Sides of square] and OR = AP = r [Sides of square]
Now, BP = AB – AP = 6 – r and, CR = AC – AR = 8 – r
Since tangents from an external point are equal
CR = CQ = 8 – r and BP = BQ = 6 – r
Now, In ΔABC,
BC^{2} = AC^{2} + AB^{2}⇒ (CQ + BQ)^{2 }= (8)^{2} + (6)^{2}⇒ (8 – r + 6 – r)^{2} = 64 + 36
⇒ (14 – 2r)^{2} = 100
⇒ (14 – 2r)^{2} = (10)^{2}⇒ 14 – 2r = 10
⇒ –2r = –4
⇒ r = 2
Hence radius of circle (r) = 2 cm.