The length of the common chord of two intersecting circles is 12 cm. If the diameters of the circles are 15 cm and 13 cm, then what is the distance (in cm) between their centers?

Given : Radius OA = $$\frac{15}{2}$$ = 7.5 cm and O'A = 6.5 cm and AB = 12 cm

To find : OO' = ?

Solution : AC = BC = $$\frac{1}{2}(AB)=\frac{12}{2}=6$$ cm

In $$\triangle$$ OAC,

=> $$(OC)^2=(OA)^2-(AC)^2$$

=> $$(OC)^2=(7.5)^2-(6)^2$$

=> $$(OC)^2=56.25-36=20.25$$

=> $$OC=\sqrt{20.25}=4.5$$ cm

Similarly, $$O'C=\sqrt{(6.5)^2-(6)^2}=\sqrt{42.25-36}$$

=> $$O'C=\sqrt{6.25}=2.5$$ cm

$$\therefore$$ Distance between the centres = OO' = OC + O'C

= $$4.5+2.5=7$$ cm

=> Ans - (B)