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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the equation of circle passing through (1,1) and the point of intersection of the circles $x^2+y^2+10x-y=0$ and $2x^2+2y^2+2x-5y-23=0$

$\begin{array}{1 1}(a)\;(x^2-y^2-10x-y)+\large\frac{22}{23}\normalsize(x^2+y^2+x-3y-\large\frac{23}{2})\normalsize=0\\(b)\;(x^2+y^2+10x-y)+\large\frac{22}{23}\normalsize(x^2+y^2+x-3y-\large\frac{23}{2})\normalsize=0\\(c)\;(x^2-y^2-10x-y)+\large\frac{23}{22}\normalsize(x^2+y^2+x-3y-\large\frac{23}{2})\normalsize=0\\(d)\;(x^2-y^2-10x-y)+\large\frac{12}{23}\normalsize(x^2+y^2+x-3y-\large\frac{23}{2})\normalsize=0\end{array}$

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Given circles are $x^2+y^2+10x-y=0$ and $2x^2+2y^2+2x-6y-23=0$
$x^2+y^2+x-\large\frac{6}{2}$$y-\large\frac{23}{2}$$=0$
equation of circle passing through the point of intersection of circle is
$(x^2+y^2+10x-y)+\lambda(x^2+y^2+x-\large\frac{6}{2}$$y-\large\frac{23}{2})$$=0$
Since it pass through (1,1) hence
$11+\lambda(-\large\frac{23}{2})$$=0$
$11=\lambda \times \large\frac{23}{2}$
$\lambda=\large\frac{22}{23}$
Hence equation of circle is
$(x^2+y^2+10x-y)+\large\frac{22}{23}\normalsize(x^2+y^2+x-3y-\large\frac{23}{2})\normalsize=0$
Hence (b) is the correct answer.
answered Feb 12, 2014 by sreemathi.v
 

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