Let $6$ and $8$ be the $X$ and $Y$-intercepts made by the circle $S \equiv x^2+y^2+2gx+2fy+c=0$ respectively. If $gx+fy+1=0$ is a line passing through the point $(1, -1)$,then the radius of the circle $S=0$ is

The equation of the circle which touches the coordinate axes and the line $\frac{x}{3} + \frac{y}{4} = 1$,and whose centre lies in the first quadrant,is ${x^2} + {y^2} - 2cx - 2cy + {c^2} = 0$,where $c$ is:

If $(6, -k)$ and $(-3, 2)$ are conjugate points with respect to the circle $x^2 + y^2 + 6x + 4y + 12 = 0$,then $k$ equals:

The area of the circle passing through the points $(5, 2), (5, -2),$ and $(1, 2)$ is (in $\pi$)

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 .... .

The length of the intercept that the circle $x^2 + y^2 + 10x - 6y + 9 = 0$ makes on the $x$-axis is: