Question

Prove that the radii of the circles x² + y² = 1, x² + y² − 2x − 6y − 6 = 0 and x² + y² − 4x − 12y − 9 = 0 are in A.P.

Answer 1

Let the radii of the circles x² + y² = 1, x² + y² − 2x − 6y − 6 = 0 and x² + y² − 4x − 12y − 9 = 0 be $r_1,r_2\mathrm{and}{r}_3$, respectively.

∴ $r_1=1,r_2=\sqrt{{\left(-1\right)}^2+{\left(-3\right)}^2+6}=4,r_3=\sqrt{{\left(-2\right)}^2+{\left(-6\right)}^2+9}=7$

Now, $r_2-r_1=r_3-r_2=3$

∴ $r_1,r_2\mathrm{and}{r}_3$ are in A.P.