The condition that the circle ${(x - 3)^2} + {(y - 4)^2} = {r^2}$ lies entirely within the circle ${x^2} + {y^2} = {R^2},$ is
$R + r \le 7$
${R^2} + {r^2} < 49$
${R^2} - {r^2} < 25$
$R - r > 5$
Circles ${x^2} + {y^2} + 2gx + 2fy = 0$ and ${x^2} + {y^2}$ $ + 2g'x + 2f'y = $ $0$ touch externally, if
Two tangents are drawn from a point $P$ on radical axis to the two circles touching at $Q$ and $R$ respectively then triangle formed by joining $PQR$ is
The equation of radical axis of the circles ${x^2} + {y^2} + x - y + 2 = 0$ and $3{x^2} + 3{y^2} - 4x - 12 = 0,$ is
If the circles ${x^2}\, + {y^2}\, - 16x\, - 20y\, + \,164\,\, = \,\,{r^2}$ and ${(x - 4)^2} + {(y - 7)^2} = 36$ intersect at two distinct points, then
If a circle $C$ passing through $(4, 0)$ touches the circle $x^2 + y^2 + 4x - 6y - 12 = 0$ externally at a point $(1, -1),$ then the radius of the circle $C$ is