Question

Two circles of radii 10 cm and 8 cm intersects each other and the length of the common chord is 12 cm find the distance between their centers. (Fig.4.36)

Answer 1

Consider the intersecting point of OP and QR as M.

$$\begin{gathered} \text{W}e\mathrm{know}\mathrm{that}\text{if}\mathrm{two}\mathrm{circles}\mathrm{intersect}\text{at}\mathrm{two}\mathrm{points},\mathrm{then}\mathrm{the}\mathrm{line}\mathrm{passing}\mathrm{through}\mathrm{their}\mathrm{centre}\mathrm{is}\mathrm{the}\mathrm{perpendicular}\mathrm{bisector} \\ \mathrm{of}\mathrm{the}\mathrm{common}\mathrm{chord}. \\ \mathrm{So},\mathrm{QM}=\mathrm{MR} \\ \mathrm{And}\mathrm{QR}\hspace{0.17em}=12\mathrm{cm} \\ \mathrm{Therefore},\mathrm{QM}=\frac{12}{2}=6\mathrm{cm}. \\ \mathrm{In}\mathrm{right}△\mathrm{QMO}, \\ \mathrm{OM}=\sqrt{{\mathrm{OQ}}^2-{\mathrm{QM}}^2} \\ =\sqrt{{10}^2-6^2} \\ =\sqrt{100-36} \\ =\sqrt{64} \\ =8\mathrm{cm} \\ \mathrm{And}\mathrm{in}\mathrm{right}△\mathrm{QMP}, \\ \mathrm{MP}=\sqrt{{\mathrm{PQ}}^2-{\mathrm{QM}}^2} \\ =\sqrt{8^2-6^2} \\ =\sqrt{28} \\ =2\sqrt{7}\mathrm{cm} \\ \therefore \mathrm{OP}=\mathrm{OM}+\mathrm{MP}=\left(8+2\sqrt{7}\right)\mathrm{cm}. \end{gathered}$$