The equation of the circle through the points | vedclass

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Question

(c) 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 +

Vedclass · Accepted Answer

${x^2} + {y^2} - x - 2y = 0$

Vedclass · Answer

(c) Family of circles is

${x^2} + {y^2} - 2x - 4y + 1 + \lambda ({x^2} + {y^2} - 1) = 0$.

$(1 + \lambda ){x^2} + (1 + \lambda ){y^2} - 2x - 4y + (1 - \lambda ) = 0$

${x^2} + {y^2} - \frac{2}{{1 + \lambda }}x - \frac{4}{{1 + \lambda }}y + \frac{{1 - \lambda }}{{1 + \lambda }} = 0$.....(i)

Centre is $\left[ {\frac{1}{{1 + \lambda }},\;\frac{2}{{1 + \lambda }}} \right]$
and radius $ = \sqrt {{{\left( {\frac{1}{{1 + \lambda }}} \right)}^2} + {{\left( {\frac{2}{{1 + \lambda }}} \right)}^2} - \frac{{1 - \lambda }}{{1 + \lambda }}} $

$= \frac{{\sqrt {4 + {\lambda ^2}} }}{{1 + \lambda }}$.

Since it touches the line $x + 2y = 0$, hence

Radius = Perpendicular from centre to the line

$i.e.$, $\left| {\frac{{\frac{1}{{1 + \lambda }} + 2\frac{2}{{1 + \lambda }}}}{{\sqrt {{1^2} + {2^2}} }}} \right| = \frac{{\sqrt {4 + {\lambda ^2}} }}{{1 + \lambda }} $

$\Rightarrow \sqrt 5 = \sqrt {4 + {\lambda ^2}} $

$\Rightarrow \lambda \pm 1$

$ \Rightarrow \sqrt 5 = \sqrt {4 + {\lambda ^2}}$

$\Rightarrow \lambda = \pm 1$

$\lambda = - 1$ cannot be possible in case of circle. So $\lambda = 1$.

Thus, from (i) ${x^2} + {y^2} - x - 2y = 0$ is the required equation of the circle.

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

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