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.
BT2 = CT × AT
⇒ 144 cm2 = CT × 13 cm
⇒ CT = 144/13 cm
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