The equation of the circle through the points of intersection of \(x^2 + y^2 – 1\) = 0, \(x^2 + y^2 – 2x – 4y + 1\) = 0 and touching the line x + 2y = 0, is

Solution :

Family of circles is \(x^2 + y^2 – 2x – 4y + 1\) + \(\lambda\)(\(x^2 + y^2 – 1\)) = 0

(1 + \(\lambda\))\(x^2\) + (1 + \(\lambda\))\(y^2\) – 2x – 4y + (1 – \(\lambda\))) = 0

\(x^2 + y^2 – {2\over {1 + \lambda}} x – {4\over {1 + \lambda}}y + {{1 – \lambda}\over {1 + \lambda}}\) = 0

Centre is (\({1\over {1 + \lambda}}\), \({2\over {1 + \lambda}}\))  and radius = \(\sqrt{4 + {\lambda}^2}\over |1 + \lambda|\)

Since it touches the line x + 2y = 0, hence

Radius = Perpendicular distance from center to the line

i.e., |\({1\over {1 + \lambda}} + 2{2\over {1 + \lambda}}\over \sqrt{1^2 + 2^2}\)| = \(\sqrt{4 + {\lambda}^2}\over |1 + \lambda|\) \(\implies\) \(\sqrt{5}\) = \(\sqrt{4 + {\lambda}^2}\) \(\implies\) \(\lambda\) = \(\pm\) 1.

\(\lambda\) = -1 cannot be possible in case of circle. So \(\lambda\) = 1.

Hence the equation of the circle is \(x^2 + y^2 – x – 2y\) = 0


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