# 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. - Mathematics

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.

#### Solution

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.

Concept: Circle - Standard Equation of a Circle
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 24 The circle
Exercise 24.2 | Q 9 | Page 32