If one common tangent of the two circles $x^2 + y^2 = 4$ and ${x^2} + {\left( {y - 3} \right)^2} = \lambda ,\lambda  > 0$ passes through the point $\left( {\sqrt 3 ,1} \right)$, then possible value of  $\lambda$ is

Let $C_1, C_2$ be two circles touching each other externally at the point $A$ and let $A B$ be the diameter of circle $C_1$. Draw a secant $B A_3$ to circle $C_2$, intersecting circle $C_1$ at a point $A_1(\neq A)$, and circle $C_2$ at points $A_2$ and $A_3$. If $B A_1=2, B A_2=3$ and $B A_3=4$, then the radii of circles $C_1$ and $C_2$ are respectively

Two circles with equal radii intersecting at the points $(0, 1)$ and $(0, -1).$ The tangent at the point $(0, 1)$ to one of the circles passes through the centre of the other circle. Then the distance between the centres of these circles is

Let the mirror image of a circle $c_{1}: x^{2}+y^{2}-2 x-$ $6 y+\alpha=0$ in line $y=x+1$ be $c_{2}: 5 x^{2}+5 y^{2}+10 g x$ $+10 f y +38=0$. If $r$ is the radius of circle $c _{2}$, then $\alpha+6 r^{2}$ is equal to$.....$

If the curves, $x^{2}-6 x+y^{2}+8=0$ and $\mathrm{x}^{2}-8 \mathrm{y}+\mathrm{y}^{2}+16-\mathrm{k}=0,(\mathrm{k}>0)$ touch each other at a point, then the largest value of $\mathrm{k}$ is

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