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 ?

50 cm

AC = Length of the direct common tangents

BD = Length of direct transverse tangents

Let, the distance between two circles = x cm

So, BD = √(\( { x}^{2 } \) - \( { 7\; +\; 7}^{2 } \))

⇒ 48 = √(\( { x}^{2 } \) - \( { 14}^{2 } \))

 Squaring both sides

⇒ \( { 48}^{2 } \) = \( { x}^{2 } \) - 196

⇒ 2304 = \( { x}^{2 } \) - 196

⇒ \( { x}^{2 } \) = 2304 + 196 = 2500

⇒ x = \(\sqrt {2500 }\) =  50 cm

ALso AC = √(\( { 50}^{2 } \) - \( { 7\; -\; 7}^{2 } \))

⇒ AC = \(\sqrt {2500 }\) = 50 cm

Therefore, the length of BD is 48 cm and AC is 50 cm.