Franco Avila

2023-03-11

How to find the center and radius of the circle ${x}^{2}+{y}^{2}+4x-8y+4=0$?

loise9oa

The $\text{general form of the equation of a circle}$ is
$\mid \overline{\underline{\frac{a}{a}{x}^{2}+{y}^{2}+2gx+2fy+c=0\frac{a}{a}\mid }}$
with centre = $\mid \overline{\underline{\frac{a}{a}\left(\begin{array}{cc}-g& -f\end{array}\right)\frac{a}{a}\mid }}$
and radius = $\mid \overline{\underline{\frac{a}{a}\sqrt{{g}^{2}+{f}^{2}-c}\frac{a}{a}\mid }}$
${x}^{2}+{y}^{2}+4x-8y+4=0\phantom{\rule{1ex}{0ex}}\text{is in this form}$
We must first determine g, f, and c in order to determine the center and radius.
By comparing the coefficients of 'like terms' in the given equation with the general form.
2g = 4 → g = 2 , 2f = -8 → f = -4 and c = 4
$⇒\phantom{\rule{1ex}{0ex}}\text{centre}=\left(-g,-f\right)=\left(-2,4\right)$
and radius $=\sqrt{{2}^{2}+{\left(-4\right)}^{2}-4}=\sqrt{4+16-4}=4$
Alternatively Use the method of $\text{completing the square}$
Add ${\left(\frac{1}{2}\text{coefficient of x/y terms}\right)}^{2}\phantom{\rule{1ex}{0ex}}\text{to both sides}$
$\left({x}^{2}+4x\text{+4}\right)+\left({y}^{2}-8y\text{+16}\right)=\text{4+16}-4$
$⇒{\left(x+2\right)}^{2}+{\left(y-4\right)}^{2}=16$
The $\text{standard form of the equation of a circle}$ is
$\mid \overline{\underline{\frac{a}{a}{\left(x-a\right)}^{2}+{\left(y-b\right)}^{2}={r}^{2}\frac{a}{a}\mid }}$
where (a ,b) are the coordinates of the centre and r, the radius.
By comparison of equation with standard form.
a = -2 , b = 4 and r = 4
Thus centre = (-2 ,4) and radius = 4

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