NIMCET 2021 Question Paper with solutions

Given Circle, ${\mathit{x}}^2+{\mathit{y}}^2=4$

Centre of first circle is $\left(0,0\right)$
Centre of second circle is $\left(3,4\right)$
Distance between centre of two circle $\left(0,0\right)$ to $\left(3,4\right)=\sqrt{{\left(3-0\right)}^2+{\left(4-0\right)}^2}$
$\sqrt{9+16}=\sqrt{25}=5$
Radius of first circle is 2
and radius of second circle is $\sqrt{{\left(3\right)}^2+{\left(4\right)}^2+24}$
$=\sqrt{25+24}$
$=\sqrt{49}=7$
${\mathit{R}}_2-{\mathit{R}}_1={\mathit{C}}_1{\mathit{C}}_2$
$7-2=5$Where ${\mathit{R}}_2=$ Radius of Second Circle
${\mathit{R}}_1=$ Radius of First Circle
${\mathit{C}}_1{\mathit{C}}_2=$ Distance between centre
Here first circle touch second circle internally
Hence, there is only one common tangent.