Question Number 115830 by mohammad17 last updated on 28/Sep/20
$${create}\:{the}\:{differention}\:{equation}\:{from} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{ax}+\mathrm{2}{by}+{c}=\mathrm{0} \\ $$
Commented by bemath last updated on 29/Sep/20
![⇒x^2 +2ax+a^2 +y^2 +2by=a^2 +b^2 −c ⇒(x+a)^2 +(y+b)^2 = a^2 +b^2 −c ⇒(d/dx) [ (x+a)^2 +(y+b)^2 ] = 0 2(x+a) + 2(y+b)y′ = 0 ⇒y′ = −(((x+a))/((y+b)))](https://www.tinkutara.com/question/Q115873.png)
$$\Rightarrow{x}^{\mathrm{2}} +\mathrm{2}{ax}+{a}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{by}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c} \\ $$$$\Rightarrow\left({x}+{a}\right)^{\mathrm{2}} +\left({y}+{b}\right)^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c} \\ $$$$\Rightarrow\frac{{d}}{{dx}}\:\left[\:\left({x}+{a}\right)^{\mathrm{2}} +\left({y}+{b}\right)^{\mathrm{2}} \:\right]\:=\:\mathrm{0} \\ $$$$\mathrm{2}\left({x}+{a}\right)\:+\:\mathrm{2}\left({y}+{b}\right){y}'\:=\:\mathrm{0} \\ $$$$\:\Rightarrow{y}'\:=\:−\frac{\left({x}+{a}\right)}{\left({y}+{b}\right)} \\ $$