Q.Find the equation of the circle which passes through the point $(2, 0)$, $(0, 2)$ and orthogonal to the circle $2x^2+2y^2+5x-6y+4=0$.

Set up the general circle through the two given points, then impose the orthogonality condition $2g_1g_2+2f_1f_2=c_1+c_2$ against the given circle.

Let the required circle be $x^2+y^2+2gx+2fy+c=0$.

Through $(2,0)$: $4+4g+c=0 \Rightarrow 4g+c=-4$ ... (i)

Through $(0,2)$: $4+4f+c=0 \Rightarrow 4f+c=-4$ ... (ii)

Subtracting, $4g=4f \Rightarrow g=f$.

The given circle $2x^2+2y^2+5x-6y+4=0$, written with unit leading coefficient, is $x^2+y^2+\frac52x-3y+2=0$, so $g_2=\frac54$, $f_2=-\frac32$, $c_2=2$.

Orthogonality condition: $2gg_2+2ff_2=c+c_2$

$2g\cdot\frac54+2g\cdot\left(-\frac32\right)=c+2$ (using $f=g$)

$\frac52g-3g=c+2 \Rightarrow -\frac12g=c+2 \Rightarrow g=-2c-4$

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