# AB is a diameter of a circle. The length of AB=5cm. If O is the centre of the circle and the length of tangent segment BT=12cm , determime CT ? Since AB is the diameter and PB is the tangent.

So, AB ⊥ PB  (Radius of the circle is perpendicular to the tangent through point of contacy)

⇒ ABC is a right triangle right angled at B

⇒ AT2 = AB2 + BT2 = (5 cm)2 + (12 cm)2 = 25 cm2 + 144 cm2 = 169 cm2  (Pythagoras theorem)

⇒ AT = 13 cm

Now, we know that

If PAB is a secant to a circle intersecting the circle to A and B and PT is a tangent, then PA × PB = PT2

Here is the link for the same.

https://www.meritnation.com/ask-answer/question/dear-meritnation-experts-plz-answer-my-question-if-pab-i/circles/1726596

So,

BT2 = CT × AT

⇒ 144 cm2 = CT × 13 cm

⇒ CT = 144/13 cm

• 9

itz 13 cm

• -9

Given- AB is a diameter, AB=5cm and BT=12cm , Bt is a tangent

To Find- CT = ?

Construction - Join BC

Solution- In triangle ABT by pythagores theorem AT = 13 ( ABT = 90 degrees, Tangent perpendicular to radius)

Let CT be x therfore AC is 13-x

Angle ABC=90 degrees  (Angle in a semicircle)

Therefore by pythagores theorem

AB^2-AC^2=CB^2 - (A)

BT^2-CT^2=CB^2 - (B)

From A and B -->

25 - (13-x)^2 = 144 - x^2

25 - 169 +26x = 144 (x^2 on both sides cancels out)

Therefore CT = x =144/13 = 11.07 cm

• 28

Ronald we have to find CT not AT

• -3

Oooops ... sorry I got it wrong thought itz for AT .

• -9
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