The equation of the circle which cuts the circle $x^2+y^2+2x+4y-4=0\quad$ and the lines $xy-2x-y+2=0\quad$ orthogonally, is
$a.\quad x^2+y^2-2x-4y-6=0\;$
$b.\quad x^2+y^2-2x-4y+6=0\;$
$c.\quad x^2+y^2-2x-4y+12=0\;$
$d.\quad$ (not possible to determine.)
Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0\quad$. As this is intersecting the circle $x^2+y^2+2x+4y-4=0\quad$ orthogonally, so $2g+4f=c-4\;$ by using the condition of orthogonality of Terri circles $2g_1g_2+2f_1f_2=c_1+c_2\;$ The second equation is an equation of pair of lines,$(x-1)(y-2)=0\quad$ But I dont know what is the condition of orthogonality of a circle and a pair of lines. I am unable to solve it further. Please help me.

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    $\begingroup$ Hint: What can one say about the location of the centre O of a circle C if C meets a line L orthogonally? $\endgroup$ – Did Oct 23 '15 at 9:05

For two figures to be orthogonal vis-à-vis each other means that at each of their points of intersection their slopes are perpendicular. In terms of analytic geometry, this would mean that the slope of one is the negative reciprocal of the slope of the other -- i.e., the product of the two slopes at the intersection equals $-1\;$.

$xy-2x-y+2=0\;$ is really the two lines $x=1$ and $y=2\;$. Any circle orthogonal to the two lines must therefore be centered at $A(1\mid 2)\;$ .
The two tangents from $A\;$ to the circle $B\equiv[x^2+y^2+2x+4y-4=0]\;$ can be constructed by drawing the circle whose diameter is the line joining the center of the given circle $C(-1\mid -2)\;$ to $A\;$. The equation of this new circle is $D\equiv[x^2+y^2=5]\;$. The two intersections $E_1\left(\frac{-1-6\sqrt{11}}{10}\mid\frac{-2+3\sqrt{11}}{10}\right)\;$ and $E_2\left(\frac{-1+6\sqrt{11}}{10}\mid\frac{-2-3\sqrt{11}}{10}\right)\;$ of the two circles $B\;$ and $D\;$ are the points of tangency.
Your required circle $[(x-1)^2+(y-2)^2=11]\;$, centered, as required, at $A\;$, passes through $E_1\;$ and $E_2\;$.

Oh, by the way, from the menu of choices you provided, the proper choice is $a.$


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