# How do you find the equation of the circle given center at (1,-3), and radius 10?

May 24, 2016

$100 = {\left(x - 1\right)}^{2} {\left(y + 3\right)}^{2}$

#### Explanation:

Suppose the centre of the circle was at the origin (where the x axis crosses the y axis). Then the equation would be:

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

The reason for this format is that the length of the radius (which is of fixed length) can be related to x and y by Pythagoras. However, the circle centre is not at the origin. It is at

$\left(x , y\right) \to \left(1 , - 3\right)$

So we can mathematically make this work by 'theoretically' moving the actual centre to a new centre located at the origin.

Thus we would have:

${r}^{2} = {\left(x - 1\right)}^{2} + {\left(y - \left(- 3\right)\right)}^{2}$

${r}^{2} = {\left(x - 1\right)}^{2} + {\left(y + 3\right)}^{2}$

But the radius is 10 so we have

${\left(10\right)}^{2} = {\left(x - 1\right)}^{2} + {\left(y + 3\right)}^{2}$

$100 = {\left(x - 1\right)}^{2} + {\left(y + 3\right)}^{2}$