﻿ 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. from Mathematics Circles Class 10 Uttarakhand Board
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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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

= 256 + 105
= 361

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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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# 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°
∠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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