If $C(\alpha, \beta)$ with $\alpha < 0$ is the centre of the circle that touches the $Y$-axis at $(0, 3)$ and makes an intercept of length $2$ units on the positive $X$-axis,then $(\alpha, \beta) =$

The equation of the circle whose centre is $(3, -1)$ and which cuts off a chord of length $6$ on the line $2x - 5y + 18 = 0$ is

The position of the point $(1, 1)$ with respect to the circle $x^2 + y^2 - x + y - 1 = 0$ is:

The circle $x^2 + y^2 + 4x - 4y + 4 = 0$ touches

The condition that the circle $(x - 3)^2 + (y - 4)^2 = r^2$ lies entirely within the circle $x^2 + y^2 = R^2$ is:

Let the circles $C_1: (x-\alpha)^2 + (y-\beta)^2 = r_1^2$ and $C_2: (x-8)^2 + (y-\frac{15}{2})^2 = r_2^2$ 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 $(\alpha+\beta) + 4(r_1^2 + r_2^2)$ equals