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


Similar Questions

Find the equation of circle having the lines \(x^2\) + 2xy + 3x + 6y = 0 as its normal and having size just sufficient to contain the circle x(x – 4) + y(y – 3) = 0.

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

If the lines 3x-4y-7=0 and 2x-3y-5=0 are two diameters of a circle of area 49π square units, then what is the equation of the circle?

The length of the diameter of the circle which touches the X-axis at the point (1,0) and passes through the point (2,3) is

The equation of the circle passing through the foci of the ellipse \(x^2\over 16\) + \(y^2\over 9\) = 1 and having center at (0, 3) is

Leave a Comment

Your email address will not be published. Required fields are marked *

Ezoicreport this ad