Question Details

Two common tangents AC and BD touch two equal circles equal of radius 7 cm, at points A, C, B and D, respectively, as shown in the figure . If the length of BD is 48 cm, what is the length of AC?

Options

40 cm

30 cm

50 cm

48 cm

Show Answer

Correct Answer :

Option C

50 cm

Solution :

The correct answer is 50 cm.

Step-by-step Explanation:

In the given figure, we have two equal circles of radius $r = 7 cm$ . Let their centers be $O_{1}$ and $O_{2}$ , and the distance between their centers be $d$ .

There are two types of tangents shown:
1. $A C$ is a direct common tangent touching the circles at points $A$ and $C$ .
2. $B D$ is a transverse common tangent touching the circles at points $B$ and $D$ .

The formula for the length of a transverse common tangent $B D$ is given by:
$B D = \sqrt{d^{2} - {(r_{1} + r_{2})}^{2}}$
Since the two circles are equal in radius:
$r_{1} = r_{2} = r = 7 cm$
Thus, the sum of the radii is:
$r_{1} + r_{2} = 7 + 7 = 14 cm$

We are given that the length of the transverse common tangent $B D = 48 cm$ . Substituting the values into the formula:
$48 = \sqrt{d^{2} - 14^{2}}$
Squaring both sides of the equation:
$48^{2} = d^{2} - 14^{2}$
$2304 = d^{2} - 196$
Solving for $d^{2}$ :
$d^{2} = 2304 + 196$
$d^{2} = 2500$
Taking the square root:
$d = \sqrt{2500} = 50 cm$

The formula for the length of a direct common tangent $A C$ is given by:
$A C = \sqrt{d^{2} - {(r_{1} - r_{2})}^{2}}$
Since the circles are equal in radius ( $r_{1} = r_{2}$ ), the term ${(r_{1} - r_{2})}^{2}$ becomes zero:
$A C = \sqrt{d^{2} - 0} = d$
Therefore, the length of the direct common tangent $A C$ is equal to the distance between the centers:
$A C = 50 cm$