Question
Two spheres of radii $$3$$ and $$4$$ cut orthogonally. The radius of common circle is :
A.
$$12$$
B.
$$\frac{{12}}{5}$$
C.
$$\frac{{\sqrt {12} }}{5}$$
D.
$${\sqrt {12} }$$
Answer :
$$\frac{{12}}{5}$$
Solution :
For the orthogonal section $${C_1}P$$ and $${C_2}P$$ are pendicular where $${C_1}$$ and $${C_2}$$ are centres of sphere of radii $$4$$ and $$3$$ respectively.
Now $${C_1}P = 4$$ and $${C_2}P = 3,$$ so $$\tan \,\theta = \frac{3}{4}$$
$$\therefore $$ Radius of circle of intersection
$$OP = {C_1}P\,\sin \,\theta = 4 \times \frac{3}{5} = \frac{{12}}{5}$$
For the orthogonal section $${C_1}P$$ and $${C_2}P$$ are pendicular where $${C_1}$$ and $${C_2}$$ are centres of sphere of radii $$4$$ and $$3$$ respectively.
Now $${C_1}P = 4$$ and $${C_2}P = 3,$$ so $$\tan \,\theta = \frac{3}{4}$$
$$\therefore $$ Radius of circle of intersection
$$OP = {C_1}P\,\sin \,\theta = 4 \times \frac{3}{5} = \frac{{12}}{5}$$