Question

o is the center of the circle with radius 5cm. p is a point at the distance 13 from the center of the circle. AB is another tangent to the circle which touches the circle at E. calculate the length of AB​

Answer 1

Step-by-step explanation:

(1) It is given that line AB is tangent to the circle at A.

∴ ∠CAB = 90º (Tangent at any point of a circle is perpendicular to the radius throught the point of contact)

Thus, the measure of ∠CAB is 90º.

i(2) Distance of point C from AB = 6 cm (Radius of the circle)

(3) ∆ABC is a right triangle.

CA = 6 cm and AB = 6 cm

Using Pythagoras theorem, we have

BC2=AB2+CA2⇒BC=62+62−−−−−−√ ⇒BC=62–√ cm

Thus, d(B, C) = 62–√ cm

(4) In right ∆ABC, AB = CA = 6 cm

∴ ∠ACB = ∠ABC (Equal sides have equal angles opposite to them)

Also, ∠ACB + ∠ABC = 90º (Using angle sum property of triangle)

∴ 2∠ABC = 90º

⇒ ∠ABC = 90°2 = 45º

Thus, the measure of ∠ABC is 45º.

Answer 2

Angle OPT =90°

In angle OPT by pythagoras theorm

OT^2= OP^2 + PT^2

13^2 =5 ^2 +PT^2

PT=12cm

Since length of tangents draw from a point to a cirlce are equal.

AP=AE=x(say)

AT=PT- AP=12 - x

Since AB is a tangent to the circle E

OE | AB

angle OEA= 90°

angle AET= 90°

AT^2=AE^2+ET^2

(12 -x)^2= x^2+(13 -5 )^2

x = 10/3 cm

Similarly,BE =10/3

AB=AE+BE=(10/3 +10/3)cm=20/7 cm

Overall ans..20/7cm

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