Let the mirror image of a circle $c_{1}: x^{2}+y^{2}-2x-6y+\alpha=0$ in the line $y=x+1$ be $c_{2}: 5x^{2}+5y^{2}+10gx+10fy+38=0$. If $r$ is the radius of circle $c_{2}$,then $\alpha+6r^{2}$ is equal to:

Let $S$ be the set of all points $\left(\frac{a}{b}, \frac{c}{d}\right)$ on the circle with radius $1$ centered at $(0,0)$,where $a$ and $b$ are relatively prime integers,$c$ and $d$ are relatively prime integers (that is,$\operatorname{HCF}(a, b) = \operatorname{HCF}(c, d) = 1$),and the integers $b$ and $d$ are even. Then,the set $S$ is:

The length of the diameter of the circle which touches the $x-$axis at the point $(1,0)$ and passes through the point $(2,3)$ is:

$A$ point inside the circle $x^2 + y^2 + 3x - 3y + 2 = 0$ is

If the straight lines $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$ are the tangents to the same circle,then the area of that circle (in square units) is

The equation of a circle which touches both axes and the line $3x - 4y + 8 = 0$ and whose centre lies in the third quadrant is