The equation of common tangent to the circles $\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}+10 \mathrm{y}+20=0$…
The equation of common tangent to the circles $\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}+10 \mathrm{y}+20=0$ and $\mathrm{x}^2+\mathrm{y}^2+8 \mathrm{x}-6 \mathrm{y}-24=0$ is
$3 x-4 y+11=0$
$3 x-4 y-11=0$
$-3 x-4 y+11=0$
$3 x+4 y+11=0$
Solution
Circle $x^2+y^2-4 x+10 y+20=0$ has centre $C_1=(2,-5)$ and radius
$\mathrm{r}_1=\sqrt{4+25-20}=3$
Circle $\mathrm{x}^2+\mathrm{y}^2+8 \mathrm{x}-6 \mathrm{y}-24=0$ has centre $\mathrm{C}_2=(2,-5)$ and radius
$\mathrm{r}_2=\sqrt{16+9+24}=7$
Distance between centres
$=\sqrt{(2+4)^2+(5-2)^2}=10$
Thus circle touch each other externally at one point only.
$\therefore$ Equation of common tangent is
$\begin{aligned}
& \left(x^2+y^2-4 x+10 y+20\right)-\left(x^2+y^2+8 x-6 y-24\right)=0 \text { i.e. } \\
& 12 x-16 y-44=0 \quad \Rightarrow \quad 3 x-4 y-11=0
\end{aligned}$