Franco Avila

2023-03-11

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

loise9oa

Beginner2023-03-12Added 4 answers

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

$\Rightarrow \phantom{\rule{1ex}{0ex}}\text{centre}=(-g,-f)=(-2,4)$

and radius $=\sqrt{{2}^{2}+{(-4)}^{2}-4}=\sqrt{4+16-4}=4$

Alternatively Use the method of $\text{completing the square}$

Add $\left(\frac{1}{2}\text{coefficient of x/y terms}\right)}^{2}\phantom{\rule{1ex}{0ex}}\text{to both sides$

$({x}^{2}+4x{\text{+4}})+({y}^{2}-8y{\text{+16}})={\text{4+16}}-4$

$\Rightarrow {(x+2)}^{2}+{(y-4)}^{2}=16$

The $\text{standard form of the equation of a circle}$ is

$\mid \overline{\underline{{\frac{a}{a}}{{(x-a)}^{2}+{(y-b)}^{2}={r}^{2}}{\frac{a}{a}}\mid}}$

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

$\mid \overline{\underline{{\frac{a}{a}}{{x}^{2}+{y}^{2}+2gx+2fy+c=0}{\frac{a}{a}}\mid}}$

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

$\Rightarrow \phantom{\rule{1ex}{0ex}}\text{centre}=(-g,-f)=(-2,4)$

and radius $=\sqrt{{2}^{2}+{(-4)}^{2}-4}=\sqrt{4+16-4}=4$

Alternatively Use the method of $\text{completing the square}$

Add $\left(\frac{1}{2}\text{coefficient of x/y terms}\right)}^{2}\phantom{\rule{1ex}{0ex}}\text{to both sides$

$({x}^{2}+4x{\text{+4}})+({y}^{2}-8y{\text{+16}})={\text{4+16}}-4$

$\Rightarrow {(x+2)}^{2}+{(y-4)}^{2}=16$

The $\text{standard form of the equation of a circle}$ is

$\mid \overline{\underline{{\frac{a}{a}}{{(x-a)}^{2}+{(y-b)}^{2}={r}^{2}}{\frac{a}{a}}\mid}}$

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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