A circle C of radius 2 lies in the second quadrant and touches both the coordinate axes. Let r be the radius of a circle that has centre at the point $(2,5)$ and intersects the circle $C$ at exactly two points. If the set of all possible values of r is the interval $(\alpha, \beta)$, then $3 \beta-2 \alpha$ is equal to :

Let a circle passing through $$(2,0)$$ have its centre at the point $$(\mathrm{h}, \mathrm{k})$$. Let $$(x_{\mathrm{c}}, y_{\mathrm{c}})$$ be the point of intersection of the lines $$3 x+5 y=1$$ and $$(2+\mathrm{c}) x+5 \mathrm{c}^2 y=1$$. If $$\mathrm{h}=\lim _\limits{\mathrm{c} \rightarrow 1} x_{\mathrm{c}}$$ and $$\mathrm{k}=\lim _\limits{\mathrm{c} \rightarrow 1} y_{\mathrm{c}}$$, then the equation of the circle is :

If the image of the point $$(-4,5)$$ in the line $$x+2 y=2$$ lies on the circle $$(x+4)^2+(y-3)^2=r^2$$, then $$r$$ is equal to:

Let the circles $$C_1:(x-\alpha)^2+(y-\beta)^2=r_1^2$$ and $$C_2:(x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2$$ touch each other externally at the point $$(6,6)$$. If the point $$(6,6)$$ divides the line segment joining the centres of the circles $$C_1$$ and $$C_2$$ internally in the ratio $$2: 1$$, then $$(\alpha+\beta)+4\left(r_1^2+r_2^2\right)$$ equals