NIMCET 2021 Question Paper with solutions

Given Circle, $x^{2}+y^{2}=4$ $\& \ x^{2}+y^{2}-6 x-8 y=24$ Centre of first circle is $(0,0)$ Centre of second circle is $(3,4)$ Distance between centre of two circle $(0,0)$ to $(3,4)=\sqrt{(3-0)^{2}+(4-0)^{2}}$ $\sqrt{9+16}=\sqrt{25}=5$ Radius of first circle is 2 and radius of second circle is $\sqrt{(3)^{2}+(4)^{2}+24}$ $=\sqrt{25+24}$ $=\sqrt{49}=7$ $R_{2}-R_{1}=C_{1} C_{2}$ $7-2=5$Where $R_{2}=$ Radius of Second Circle $R_{1}=$ Radius of First Circle $C_{1} C_{2}=$ Distance between centre Here first circle touch second circle internally Hence, there is only one common tangent.