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  • The equation of the circle through the points of intersection of <img alt="\mathrm{x^2+y^2-1=0, x^2+y^2-2 x-4 y+1=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus

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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