Question

AB is diameter .TB is the tangent to the circle .If Q is a point on TB, then find the length of QA

Answer 1

Your question is not clear and appears to be incomplete.

From the given data , you can find the QA by applying Pythagoras theorem in it as ∠ABQ = 90 °.(The tangent is always perpendicular to the radius at the point of contact .)

QA  = √(AB2+QB2)

Hope this would have cleared your doubt .

Answer 2

Given:

AB is the diameter. TB is the tangent to the circle. If Q is a point on TB

To Find:

find the length of QA

Solution:

First, let us draw a diagram to visualize the given situation, draw a circle and name the radius as 'r' and then draw a tangent TB to the circle and select a point on the tangent after which we need to find the length of QA

Now we know that the diameter will be perpendicular to the tangent TB hence it will form a right angle triangle ABQ

Let the diameter be 2r and the value of QB be a

using the Pythagoras theorem we can find the value of QA, which goes as

$QA^2=(2r)^2+a^2\\QA=\sqrt{4r^2+a^}$

Using this formula whatever the value of radius be and the distance from point of contact to the point Q be we can find the value

Hence, the length of QA is $\sqrt{4r^2+a^2}$.