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 \[r_1 , r_2\ \text{and}\ r_3\], 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, 

\[r_2 - r_1 = r_3 - r_2 = 3\]

∴\[r_1 , r_2\ \text{and}\ r_3\] are in A.P.