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The equation of the circle through the points of intersection of $\mathrm{x^2+y^2-1=0, x^2+y^2-2 x-4 y+1=0}$ and touching the line $\mathrm{x+2 y=0}$ is
Option: 1
$\mathrm{x^2+y^2+x+2 y=0}$

Option: 2
$\mathrm{x^2+y^2-x+20=0}$

Option: 3
$\mathrm{x^2+y^2-x-2 y=0}$

Option: 4
$\mathrm{2\left(x^2+y^2\right)-x-2 y=0}$

Answers (1)

best_answer

Family of circles is $\mathrm{x^2+y^2-2 x-4 y+1+\lambda\left(x^2+y^2-1\right)=0}$

$\mathrm{x^2+y^2-\frac{2}{1+\lambda} x-\frac{4}{1+\lambda} y+\frac{1-\lambda}{1+\lambda}=0}$

Centre is $\mathrm{\left[\frac{1}{1+\lambda}, \frac{2}{1+\lambda}\right]}$ and radius $\mathrm{=\sqrt{\left(\frac{1}{1+\lambda}\right)^2+\left(\frac{2}{1+\lambda}\right)^2-\left(\frac{1-\lambda}{1+\lambda}\right)}=\sqrt{\frac{4+\lambda^2}{(1+\lambda)^2}}}$

Since it touches the line $\mathrm{x+2 y=0}$ Hence Radius = perpendicular distance from centre to the

line $\mathrm{=\left|\frac{\frac{1}{1+\lambda}+\frac{4}{1+\lambda}}{\sqrt{1^2+2^2}}\right|}$

$\mathrm{ =\sqrt{\frac{4+\lambda^2}{(1+\lambda)^2}}=\frac{\sqrt{4+\lambda^2}}{1+\lambda} \Rightarrow \sqrt{5}=\sqrt{4+\lambda^2} \Rightarrow \lambda= \pm 1 }$

$\mathrm{\lambda=-1}$ cannot be possible in case of circle, so $\mathrm{ \lambda=1}$ .

$\therefore$ Equation of circle is $\mathrm{x^2+y^2-x-2 y=0}$

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