If $P(a,b)$ is a point in the first quadrant.If the two circles which pass through $P$ and touch both the coordinate axes and also cut at right angles,then:
$(A)a^2-6ab+b^2=0\hspace{1cm}(B)a^2+2ab-b^2=0$
$(C)a^2-4ab+b^2=0\hspace{1cm}(D)a^2-8ab+b^2=0$
I tried to solve it.Let the equation of the circles be $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$.As the two circles are touching both the coordinate axes,so their equations are $x^2+y^2+2g_1x+2g_1y+g^2_1=0$ and $x^2+y^2+2g_2x+2g_2y+g^2_2=0$ because when circle touches both the coordinate axes,then $g=f=-r$
As the circles pass through $P(a,b)$,
so,$a^2+b^2+2g_1a+2g_1b+g^2_1=0$ and $a^2+b^2+2g_2a+2g_2b+g^2_2=0$
Then i minused the two equations to get
$2a(g_1-g_2)+2b(g_1-g_2)+(g_1-g_2)(g_1+g_2)=0$
$2a+2b+(g_1+g_2)=0$
$g_1+g_2=-(2a+2b)$
As the two circles cut orthogonally,so
$2g_1g_2+2g_1g_2=g^2_1+g^2_2$
But then i am stuck here and i cannot decide which option is correct.Please help me.Thanks.