If the two circles, $x^2 + y^2 + 2 g_1x + 2 f_1y = 0\, \& \,x^2 + y^2 + 2 g_2x + 2 f_2y = 0$ touch each then:

Let a circle $C_1 \equiv  x^2 + y^2 - 4x + 6y + 1 = 0$ and circle $C_2$ is such that it's centre is image of centre of $C_1$ about $x-$axis and radius of $C_2$ is equal to radius of $C_1$, then area of $C_1$ which is not common with $C_2$ is -

Consider the circles ${x^2} + {(y - 1)^2} = $ $9,{(x - 1)^2} + {y^2} = 25$. They are such that

If the centre of a circle which passing through the points of intersection of the circles ${x^2} + {y^2} - 6x + 2y + 4 = 0$and ${x^2} + {y^2} + 2x - 4y - 6 = 0$ is on the line $y = x$, then the equation of the circle is

The number of common tangents to the circles ${x^2} + {y^2} = 4$ and ${x^2} + {y^2} - 6x - 8y = 24$ is

If the circles $x^{2}+y^{2}+6 x+8 y+16=0$ and $x^{2}+y^{2}+2(3-\sqrt{3}) x+x+2(4-\sqrt{6}) y$ $= k +6 \sqrt{3}+8 \sqrt{6}, k >0$, touch internally at the point $P(\alpha, \beta)$, then $(\alpha+\sqrt{3})^{2}+(\beta+\sqrt{6})^{2}$ is equal to $\dots\dots$