If $x^2+y^2-4x-2y+5=0$ and $x^2+y^2-6x-4y-3=0$ are members of a coaxial system of circles,then the centre of a point circle in the system is

If $(h, k)$ is the internal centre of similitude of the circles $x^2+y^2+2x-6y+1=0$ and $x^2+y^2-4x+2y+4=0$,then $4h=$

The equation of the circle which passes through the point $(1, 1)$ and intersects the given circles $x^2 + y^2 + 2x + 4y + 6 = 0$ and $x^2 + y^2 + 4x + 6y + 2 = 0$ orthogonally is:

Find the equation of a circle with center $(4, 3)$ that touches the circle $x^2 + y^2 = 1$ internally.

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

Find the minimum radius of the circle which is orthogonal to both the circles $x^2+y^2+4x+3=0$ and $x^2+y^2-12x+35=0$.