Question

The centers of two circles with radii 8 cm and 3 cm are 13 cm apart. A direct common tangent touches the circles at A and B respectively then the length of AB is
A) 8 cm
B) 10 cm
C) 12 cm
D) None of these

Answer 1

Hint- In this question, we know that tangent to any circle makes a right angle with the radius of the circle. Also we have to make construction by drawing a line parallel to AB from ${O_2}$ to ${O_1}A$ to intersect at P as shown below.
Complete step-by-step answer:

Let us consider that ${O_1}$ and ${O_2}$ be the center of circle X and circle Y, respectively, as shown in the figure above. We have,
Radius of circle X i.e. ${O_1}A$ = 8 cm and
Radius of circle Y i.e. ${O_2}A$ = 3 cm. Also, the distance between the centers of both the circles is 13 cm$({O_1}{O_2})$.
We have the common tangent as AB on both circles. We have done the construction from ${O_2}$ to ${O_1}A$ to intersect at P as shown.
We can get the value of ${O_1}P$ as mentioned below,
\[ \Rightarrow {O_1}P = {O_1}A - {O_2}B = 8 - 3 = 5cm.\]
In $\vartriangle P{O_1}{O_2}$,
$ \Rightarrow P{O_2} = \sqrt {{{({O_1}{O_2})}^2} - {{({O_1}P)}^2}} $ (Using Pythagoras Theorem)
$ \Rightarrow P{O_2} = \sqrt {{{(13)}^2} - {{(5)}^2}} $
$ \Rightarrow P{O_2} = 12cm$. This is the same distance as that of common tangent AB.
Hence, option (C) is correct.
Note- Here we have not directly calculated the length of the common tangent. We have first calculated the length of the line which is equal to it and parallel to it. We have used Pythagoras theorem as mentioned below:
$ \Rightarrow Hypotenuse = \sqrt {{{(Perpendicular)}^2} + {{(Base)}^2}} $.