# If a circle has center (0,0) and a point on the circle (-2,-4) write the equation of the circle.?

Jun 20, 2018

${x}^{2} + {y}^{2} = 20$

#### Explanation:

The general equation of a circle of radius $r$ centred on $\left(a , b\right)$ is:

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

Thus a circle centered on the origin will have an equation of the form:

${x}^{2} + {y}^{2} = {r}^{2}$

Knowing that $\left(- 2 , 4\right)$ lies on the circle, we have:

${\left(- 2\right)}^{2} + {\left(4\right)}^{2} = {r}^{2}$

$\therefore {r}^{2} = 4 + 16 = 20$

Thus the equation is

${x}^{2} + {y}^{2} = 20$

Jun 20, 2018

${x}^{2} + {y}^{2} = 20$

#### Explanation:

Every point on a circle has the same distance from the center. This distance is the radius $r$ of the circle.

So, if $\left(- 2 , - 4\right)$ is a point on the circle, it means that the radius of the circle is the distance between $\left(0 , 0\right)$, the center, and $\left(- 2 , - 4\right)$.

To compute the distance between two points $\left({x}_{1} {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$, the formula is

$d = \setminus \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}$

In this case, $\left({x}_{1} {y}_{1}\right) = \left(0 , 0\right)$ and $\left({x}_{2} , {y}_{2}\right) = \left(- 2 , - 4\right)$. So, their distance is

$d = \setminus \sqrt{{\left(0 - \left(- 2\right)\right)}^{2} + {\left(0 - \left(- 4\right)\right)}^{2}} = \sqrt{4 + 16} = \sqrt{20}$

Now we know the center $\left(0 , 0\right)$ and the radius $\sqrt{20}$ of the circle. When you have this information, you can write the equation as

${\left(x - {x}_{0}\right)}^{2} + {\left(y - {y}_{0}\right)}^{2} = {r}^{2}$

where $\left({x}_{0} , {y}_{0}\right)$ is the center and $r$ is the radius. So, in this case, the equation is

${\left(x - 0\right)}^{2} + {\left(y - 0\right)}^{2} = {\left(\sqrt{20}\right)}^{2}$

which can be rewritten as

${x}^{2} + {y}^{2} = 20$