From three non-collinear points,we can draw:

Two circles $x^2 + y^2 = ax$ and $x^2 + y^2 = c^2$ touch each other if:

If the line $y = mx + 1$ meets the circle $x^2 + y^2 + 3x = 0$ at two points equidistant from the $y$-axis and on opposite sides,then:

$A$ circle with center at $(2,4)$ is such that the line $x+y+2=0$ cuts a chord of length $6$. The radius of the circle is

$A$ square is inscribed in the circle $x^2+y^2-2x+4y-93=0$ with its sides parallel to the coordinate axes. Which of the following can be one of the vertices of the square?

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.