Number of common tangents to the circles $x^2+y^2-6 x-14 y+48=0$ and $x^2+y^2-6 x=0$ are

Number of common tangents to the circles $x^2+y^2-6 x-14 y+48=0$ and $x^2+y^2-6 x=0$ are
  1. $0$
  2. $1$
  3. $4$
  4. $2$

Solution

$\begin{array}{ll} & x^2+y^2-6 x-14 y+48=0 \\ \therefore \quad & \mathrm{C}_1(3,7), \mathrm{r}_1=\sqrt{10} \\ & \text { Again } x^2+y^2-6 x=0 \\ \therefore \quad & \mathrm{C}_2(3,0), \mathrm{r}_2=3 \end{array}$ Now $l\left(\mathrm{C}_1 \mathrm{C}_2\right)=$ distance between centres $\begin{aligned} & \therefore \quad l\left(\mathrm{C}_1 \mathrm{C}_2\right)=\sqrt{0^2+7^2}=7 \text { and } \\ & \\ & \mathrm{r}_1+\mathrm{r}_2=\sqrt{10}+3 < l\left(\mathrm{C}_1 \mathrm{C}_2\right) \end{aligned}$ $\Rightarrow$ The given circles are disjoint. $\Rightarrow$ Number of common tangents is 4 .

Asked in: MHT CET 2023 (11 May Shift 1)

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