# How do you write the equation of the locus of all points in the coordinate plane 8 units from (4, 10)?

Jun 15, 2016

${\left(x - 4\right)}^{2} + {\left(y - 10\right)}^{2} = 64$, or ${x}^{2} + {y}^{2} - 8 x - 20 y + 52 = 0$

#### Explanation:

By definition, the locus of all the points who have the same distance from a fixed centre is a circle. So, that was a complex way to ask for a circle of center $\left(4 , 10\right)$ and radius $8$.

If you know the center coordinates $\left({x}_{0} , {y}_{0}\right)$ and the radius $r$, the formula for the cirle is given by

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

In your case, ${x}_{0} = 4$, ${y}_{0} = 10$ and $r = 8$, so the expression becomes

${\left(x - 4\right)}^{2} + {\left(y - 10\right)}^{2} = 64$

You can either leave it as it is, or expand the squares and sum all the coefficients:

${x}^{2} - 8 x + 16 + {y}^{2} - 20 y + 100 = 64$

and thus

${x}^{2} + {y}^{2} - 8 x - 20 y + 52 = 0$