Among all the parallelograms whose diagonals are $10$ and $4$,the one having maximum area has its perimeter lying in the interval

The largest among the distances from the point $P(15, 9)$ to the points on the circle $x^2 + y^2 - 6x - 8y - 11 = 0$ is

$A$ circle touches both the coordinate axes and the straight line $L \equiv 4x+3y-6=0$ in the first quadrant. If this circle lies below the line $L=0$,then the equation of that circle is

From a point $P$ on the line $4x - 3y = 6$,two tangents are drawn to the circle $x^2 + y^2 - 6x - 4y + 4 = 0$. If the angle between these tangents is $\tan^{-1}\left(\frac{24}{7}\right)$,then $P$ can be:

If $A=(0,-2)$ and $B$ is any point on the circle $x^2+y^2-2x-2y+1=0$,then the maximum value of $(AB)^2$ is

Statement $(A):$ The number of common tangents to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 6x - 8y = 24$ is $4$.
Reason $(R):$ For two circles with centers $C_1, C_2$ and radii $r_1, r_2$,if $|C_1C_2| > r_1 + r_2$,then the circles have $4$ common tangents.