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
Similar Questions
Find the equation of the normal to the circle \(x^2 + y^2 – 5x + 2y -48\) = 0 at the point (5,6).
