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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. - 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.

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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
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APPEARS IN

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