Blog: Determine the center and the radius of the circle x^2+y^2+10y=0.

Tuesday, November 4, 2014

Determine the center and the radius of the circle x^2+y^2+10y=0.

The general equation for the family of circles is

$\displaystyle{{\left({x}-{a}\right)}}^{{2}}+{{\left({y}-{b}\right)}}^{{2}}={{r}}^{{2}}$

with $\displaystyle{\left({a},{b}\right)}$ as center and $\displaystyle{r}$ radius

Let us compare our equation with the general equation

there is only one x term, so a becomes 0, i.e., $\displaystyle{{\left({x}-{a}\right)}}^{{2}}={{\left({x}-{0}\right)}}^{{2}}={{x}}^{{2}}$

but there are two terms of y. To find b value we need to find the roots of the quadratic polynomial of $\displaystyle{{y}}^{{2}}+{10}{y}$ .

One of the ways to find roots of this polynomial is completing square method.

$\displaystyle{{\left({y}\right)}}^{{2}}+{2}\cdot{y}\cdot{5}+{{\left({5}\right)}}^{{2}}-{{\left({5}\right)}}^{{2}}$ (equating this to $\displaystyle{{a}}^{{2}}+{2}{a}{b}+{{b}}^{{2}}$ identity)

$\displaystyle={{\left({y}+{5}\right)}}^{{2}}-{25}$

Now combine $\displaystyle{x}$ and $\displaystyle{y}$ terms and write the equation for the circle

$\displaystyle{{\left({x}-{0}\right)}}^{{2}}+{{\left[{y}-{\left(-{5}\right)}\right]}}^{{2}}-{25}={0}$

$\displaystyle{{\left({x}-{0}\right)}}^{{2}}+{{\left[{y}-{\left(-{5}\right)}\right]}}^{{2}}={{5}}^{{2}}$

$\displaystyle\therefore$ the center of the circle is (0,-5)

and the radius is 5

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Tuesday, November 4, 2014

Determine the center and the radius of the circle x^2+y^2+10y=0.

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