# 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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