Prove that the radii of the circles
$x^{2} + y^{2} = 1, x^{2} + y^{2} - 2 x - 6 y - 6 = 0$
and $x^{2} + y^{2} - 4 x - 12 y - 9 = 0$ are in A.P

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Solution

The given equation of circles are
$x^{2} + y^{2} = 1 \dots (i) x^{2} + y^{2} - 2 x - 6 y - 6 = 0 \dots (i i) x^{2} + y^{2} - 4 x - 12 y - 9 = 0 \dots (i i i)$
Let $R_{1} R_{2}$ and $R_{3}$ are the radii of (i) (ii) and (iii) respectively
$∴ R_{1} = 1$
$R_{2} = \sqrt{g^{2} + f^{2} + c} = \sqrt{1^{2} + 3^{2} + 6} = \sqrt{16} = 4 R_{3} = \sqrt{g^{2} + f^{2} + c} = \sqrt{2^{2} + 6^{2} + 9} = \sqrt{49} = 7$
Now $r_{2} - r_{3} - r_{2} = 3$
if a, b, c are in AP, then $b = \frac{a + c}{2}$
The centres of the three circles lie in AP.
From a=1, b = 4, c = 7
$\frac{1 + 7}{2} = 4 = b$
1, 4, 7 are in AP.

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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

Prove that the radii of the circles
$x^{2} + y^{2} = 1, x^{2} + y^{2} - 2 x - 6 y - 6 = 0$
and $x^{2} + y^{2} - 4 x - 12 y - 9 = 0$ are in A.P