﻿ From a point P, the length of the tangent to a circle is 15 cm and distance of P from the centre of the circle is 17 cm. Then what is the radius of the circle? from Mathematics Circles Class 10 Uttarakhand Board
The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle. BD is a tangent to the smaller circle touching it at D. Find the length AD. Given : Two concentric circles C1 and Cof radii 13 cm and 8 cm respectively. AB is a diameter of the bigger circle (C1) and BD is a tangent to the smaller circle (C2).
∴ 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

OB2 = OD2 = BD2                [ Using Pythagoras theorem] (13)= (8)+ BD 169 = 64 + BD2 BD2 = 169 - 64 BD2 = 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

AD2 = AE2 + DE = 256 + 105
= 361 AD  = 19   cm.

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ABCD is a quadrilateral such that ∠D = 90°. A cirlce C (O, r) touches the sides AB, BC, CD and DA at P, Q, R and S respectively. If BC = 38 cm, CD = 25 cm and BP = 27 cm, Find r.

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.

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A line intersecting a circle in two points is called a ___________.
• Diameter
• Chord

• Secant
• Secant

C.

Secant
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# From a point P, the length of the tangent to a circle is 15 cm and distance of P from the centre of the circle is 17 cm. Then what is the radius of the circle?

Since, the tangent of any point of a line is perpendicular to the radius through the point of contact. Therefore, ∠OPO =90°
It is given that PQ =15 cm
and    OQ = 17 cm
In right ΔOPQ
OQ2 = OP2 + PQ2
[Using Pythagoras theorem]
⇒ (17)2 = OP2 + (15)2
OP2 = (17)– (15)2
= 289 – 225 = 64 ⇒ OP = 8 cm
Hence, radius of the circle = 8 cm.

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In the given figure, ABC is a right-angled triangle with AB = 6 cm and AC = 8 cm. A circle with centre O has been inscribed inside the triangle. Calculate the value of r, the radius of the inscribed circle. 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
BC2 = AC2 + AB2
= (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:
∵ ∠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,
BC2 = AC2 + AB2
⇒ (CQ + BQ)= (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.

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