The two circles with radii 6 cm and 3 cm centered at the point P and Q respectively. A transverse common tangent AC to the both circles cuts line joining P and Q at the point B such that AB = 8 cm. Calculate the length (in cm) of PQ.

Given : AP = $$r_1=6$$ cm , CQ = $$r_2=3$$ cm and AB = 8 cm

To find : PQ = $$d=?$$

Solution : In right $$\triangle$$ APB right angled at A,

=> $$(PB)^2=(AP)^2+(AB)^2$$

=> $$(PB)^2=(6)^2+(8)^2$$

=> $$(PB)^2=36+64=100$$

=> $$PB=\sqrt{100}=10$$ cm --------------(i)

Similarly, in $$\triangle$$ BCQ, => $$(BQ)^2-(BC)^2=9$$ --------------(ii)

Also, AC is transverse common tangent,

=> $$(AC)^2=d^2-(r_1+r_2)^2$$

=> $$(8+BC)^2=(10+BQ)^2-(6+3)^2$$

=> $$64+(BC)^2+16(BC)=100+(BQ)^2+20(BQ)-81$$

=> $$16(BC)-20(BQ)=19-64+(BQ)^2-(BC)^2$$

Using value from equation (ii), we get :

=> $$16(BC)-20(BQ)=19-64+9$$

=> $$16(BC)-20(BQ)=-36$$

=> $$5(BQ)-4(BC)=9$$ -------------(iii)

Solving equations (ii) and (iii), we get : $$BQ = 5$$ cm and $$BC=4$$ cm

$$\therefore$$ $$PQ=10+5=15$$ cm

=> Ans - (C)