Question

In the figure PT is a tangent to the circle at T. If PA=4cm and AB=5cm then PT=

(A) 4cm
(B) 5cm
(C) 6cm
(D) $2\sqrt{5}$ cm

Answer 1

Hint: We solve this question by first assuming the radius of circle as r. Then we consider the right-angled triangle OPT and apply the Pythagoras theorem, sum of squares of the two sides of a right-angled triangle is equal to the square of the hypotenuse and we get an equation. Then we consider the triangle OSA and triangle OSP and apply the Pythagoras theorem and get two equations. Then by solving these three equations, we can find the value of length of tangent PT.

Complete step by step answer:
From the given figure, we can see that OA, OB and OT are the radius of circles. Let us assume that the radius of the circle is r.

From the above figure let us consider the triangle OTP right angled at T. Here in triangle OTP, base is PT, altitude is OT and hypotenuse is OP.
Now let us consider the formula Pythagoras Theorem, sum of squares of two sides of a right-angled triangle is equal to square of hypotenuse.
So, using it we get,
$\begin{array}{rl}& \Rightarrow O{T}^2+T{P}^2=O{P}^2\\ & \Rightarrow r^2+T{P}^2=O{P}^2\\ & \Rightarrow T{P}^2=O{P}^2-r^2.......\left(1\right)\end{array}$
Now let us consider the chord AB.
Now let us consider the property that any line perpendicular to a chord of the circle from the centre bisects the chord of the circle.
Here AB is a chord and OS is perpendicular to AB. So, from the above property we can say that S is the midpoint of AB.
So, as we have that AB=5cm, we get that $SA=SB={\displaystyle \frac{5}{2}}cm$.

Now let us consider the triangle OSA right angled at S. Here in triangle OSA, base is SA, altitude is OS and hypotenuse is OA.
Now let us consider the formula Pythagoras Theorem, sum of squares of two sides of a right-angled triangle is equal to square of hypotenuse.
So, using it we get,
$\Rightarrow O{S}^2+S{A}^2=O{A}^2$
Now substituting the value of SA from above, we get,
$\begin{array}{rl}& \Rightarrow O{S}^2+{\left({\displaystyle \frac{5}{2}}\right)}^2=r^2\\ & \Rightarrow O{S}^2+{\displaystyle \frac{25}{4}}=r^2\\ & \Rightarrow O{S}^2=r^2-{\displaystyle \frac{25}{4}}.......\left(2\right)\end{array}$
Now, let us consider the triangle OPS right angled at S. Here in triangle OSP, base is SP, altitude is OS and hypotenuse is OP.
From the above figure we can see that the side SP is,
$\begin{array}{rl}& \Rightarrow SP=SA+AP\\ & \Rightarrow SP={\displaystyle \frac{5}{2}}+4\\ & \Rightarrow SP={\displaystyle \frac{13}{2}}\end{array}$
Now let us consider the formula Pythagoras Theorem, sum of squares of two sides of a right-angled triangle is equal to square of hypotenuse.
So, using it we get,
$\begin{array}{rl}& \Rightarrow O{S}^2+S{P}^2=O{P}^2\\ & \Rightarrow O{S}^2+{\left({\displaystyle \frac{13}{2}}\right)}^2=O{P}^2\\ & \Rightarrow O{S}^2+{\displaystyle \frac{169}{4}}=O{P}^2\\ & \Rightarrow O{S}^2=O{P}^2-{\displaystyle \frac{169}{4}}.........\left(3\right)\end{array}$
Now, from equations (2) and (3), we get,
$\begin{array}{rl}& \Rightarrow O{P}^2-{\displaystyle \frac{169}{4}}=r^2-{\displaystyle \frac{25}{4}}\\ & \Rightarrow O{P}^2-r^2={\displaystyle \frac{169}{4}}-{\displaystyle \frac{25}{4}}\\ & \Rightarrow O{P}^2-r^2={\displaystyle \frac{144}{4}}\\ & \Rightarrow O{P}^2-r^2=36\end{array}$
Now, let us substitute this value in equation (1). Then we get,
$\begin{array}{rl}& \Rightarrow T{P}^2=36\\ & \Rightarrow TP=6\end{array}$

So, the correct answer is “Option C”.

Note: We can also solve this question in an alternative and simpler method.
Let us consider the Secant-Tangent Power theorem of circles.
The square of tangent is equal to the product of secant and its external segment.
Here PT is the tangent, PB is a secant and PA is its external segment.
So, by the above theorem we have,
$\begin{array}{rl}& \Rightarrow P{T}^2=PA\times PB\\ & \Rightarrow P{T}^2=PA\times (PA+AB)\end{array}$
Substituting the values given in the question, we get,
$\begin{array}{rl}& \Rightarrow P{T}^2=4\times (4+5)\\ & \Rightarrow P{T}^2=4\times 9=36\\ & \Rightarrow PT=6cm\end{array}$