Let the line $x-y+1=0$ intersect the circle $x^2+y^2+2x+2y+1=0$ at two points $A$ and $B$. If $AB$ is the diameter of the circle $x^2+y^2+2gx+2fy+c=0$,then $g+f=$

The value of $c$ for which the set,$\{(x, y) | x^2 + y^2 + 2x \le 1 \} \cap \{(x, y) | x - y + c \ge 0\}$ contains only one point in common is :

$A(x_1, y_1)$ is the internal centre of similitude and $B(x_2, y_2)$ is the external centre of similitude of two circles $C_1$ and $C_2$ whose centres are $P(\alpha, \beta)$ and $Q(\gamma, \delta)$ respectively. If $PA=3, AB=5, QB=2$,then the ratio of the radii of the two circles is:

Two circles each of radius $5$ units touch each other at $(1,2)$ and $4x+3y=10$ is their common tangent. The equation of that circle among the two given circles,such that some portion of it lies in every quadrant is

For $n \in N$,let $S_{n} = \{ z \in C : |z - 3 + 2i| = \frac{n}{4} \}$ and $T_{n} = \{ z \in C : |z - 2 + 3i| = \frac{1}{n} \}$. Then the number of elements in the set ${ n \in N : S_{n} \cap T_{n} = \phi }$ is.

If the line $ax + by = 0$ touches the circle ${x^2} + {y^2} + 2x + 4y = 0$ and is a normal to the circle ${x^2} + {y^2} - 4x + 2y - 3 = 0$,then the value of $(a, b)$ will be