If the circles $(x+1)^2+(y+2)^2=r^2$ and $x^2+y^2-4 x-4 y+4=0$ intersect at exactly two distinct points, then
$5<$ r $<9$
$0<$ r $<7$
$3<$ r $<7$
$\frac{1}{2}<$ r $<7$
The circle passing through the intersection of the circles, $x^{2}+y^{2}-6 x=0$ and $x^{2}+y^{2}-4 y=0$ having its centre on the line, $2 x-3 y+12=0$, also passes through the point
In the figure shown, radius of circle $C_1$ be $ r$ and that of $C_2$ be $\frac{r}{2}$ , where $r= \frac {1}{3} PQ,$ then length of $AB$ is (where $P$ and $Q$ being centres of $C_1$ $\&$ $C_2$ respectively)
Let $Z$ be the set of all integers,
$\mathrm{A}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbb{Z} \times \mathbb{Z}:(\mathrm{x}-2)^{2}+\mathrm{y}^{2} \leq 4\right\}$
$\mathrm{B}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbb{Z} \times \mathbb{Z}: \mathrm{x}^{2}+\mathrm{y}^{2} \leq 4\right\} \text { and }$
$\mathrm{C}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbb{Z} \times \mathbb{Z}:(\mathrm{x}-2)^{2}+(\mathrm{y}-2)^{2} \leq 4\right\}$
If the total number of relation from $\mathrm{A} \cap \mathrm{B}$ to $\mathrm{A} \cap \mathrm{C}$ is $2^{\mathrm{p}}$, then the value of $\mathrm{p}$ is :
The equation of the image of the circle ${x^2} + {y^2} + 16x - 24y + 183 = 0$ by the line mirror $4x + 7y + 13 = 0$ is
Choose the correct statement about two circles whose equations are given below
$x^{2}+y^{2}-10 x-10 y+41=0$
$x^{2}+y^{2}-22 x-10 y+137=0$