# How do you write an equation with center is (4,-1) and solution point is (1,4)?

Dec 25, 2016

I am going to assume that you want the equation of a circle. Please see the explanation.

#### Explanation:

The standard Cartesian form for the equation of a circle is:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2} \text{ [1]}$

where x and y are any point, $\left(x , y\right)$, on the circle, h and k are the center point, $\left(h , k\right)$, and r is the radius.

Substitute 4 for h and -1 for k into equation [1]:

${\left(x - 4\right)}^{2} + {\left(y - - 1\right)}^{2} = {r}^{2} \text{ [2]}$

To find the value of r, substitute 1 for x and 4 for y into equation [2], then solve for r:

${\left(1 - 4\right)}^{2} + {\left(4 - - 1\right)}^{2} = {r}^{2}$

${3}^{2} + {5}^{2} = {r}^{2}$

${r}^{2} = 36$

$r = 6$

Substitute 6 for r into equation [2]:

${\left(x - 4\right)}^{2} + {\left(y - - 1\right)}^{2} = {6}^{2} \text{ [3]}$

Equation [3] represents the desired circle.