The equation of the pair of straight lines parallel to the $x$-axis and touching the circle $x^2+y^2-6x-4y-12=0$ is

The intercept on the line $y=x$ by the circle $x^2+y^2-2x=0$ is $AB$. The equation of the circle with $AB$ as diameter is:

Three circles,each of radius $1$,touch one another externally and lie between two parallel lines. The minimum possible distance between the lines is:

If the area of an equilateral triangle inscribed in the circle $x^2 + y^2 + 10x + 12y + c = 0$ is $27\sqrt{3} \text{ sq. units}$,then $c$ is equal to

Two circles each of radius $5 \text{ units}$ touch each other at the point $(1, 2)$. If the equation of their common tangent is $4x + 3y = 10$,and $C_{1}(\alpha, \beta)$ and $C_{2}(\gamma, \delta)$,$C_{1} \neq C_{2}$ are their centres,then $|(\alpha + \beta)(\gamma + \delta)|$ is equal to .... .

Does the point $(-2.5, 3.5)$ lie inside,outside,or on the circle $x^{2}+y^{2}=25$?