# How do you graph  (x+3)^2 + (y-2)^2 = 25?

Apr 10, 2016

$x$ intercepts at: $x = \pm \sqrt{21} - 3$
$y$ intercepts at: $y = \pm 4 + 2$
Centre Point at: $\left(- 3 , 2\right)$

#### Explanation:

The function is a circular function:

The general form of a circular function can be expressed as:

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

Where the centre point of the graph is present at the point $\left(h , k\right)$
and the solution (the number after the = sign) is the radius of the circle squared.

${\left(x + 3\right)}^{2} + {\left(y - 2\right)}^{2} = 25$

We can determine that:

The centre point of the function is present at the point $\left(- 3 , 2\right)$ and the radius is $\sqrt{25} = 5$

For any function, $x$ incepts where $y$ = 0

Therefore, by substituting $y = 0$ we get:

${\left(x + 3\right)}^{2} + {\left(- 2\right)}^{2} = 25$

By simplifying this and solving for $x$ we get:

${\left(x + 3\right)}^{2} + 4 = 25$
${\left(x + 3\right)}^{2} = 25 - 4$
${\left(x + 3\right)}^{2} = 21$
$\left(x + 3\right) = \pm \sqrt{21}$
Remember that any real square has two solutions (a positive and negative), hence the $\pm \sqrt{21}$
$x = \pm \sqrt{21} - 3$
Therefore, two $x$ intercepts are present:
One at: $x = - \sqrt{21} - 3$
The other at: $x = + \sqrt{21} - 3$

For any function, $y$ intercepts where $x$ = 0

Therefore, if we substitute $x$ = 0 into your equation, we get:

${\left(0 + 3\right)}^{2} + {\left(y - 2\right)}^{2} = 25$

By simplifying and solving for $y$ we get:

$9 + {\left(y - 2\right)}^{2} = 25$
${\left(y - 2\right)}^{2} = 16$
Again, remember that any real square has two solutions (a positive and negative), hence the $\pm \sqrt{16}$
$\left(y - 2\right) = \pm \sqrt{16}$
$\left(y - 2\right) = \pm 4$
$y = \pm 4 + 2$
Therefore, two $y$ intercepts are present:
One at: $y = - 2$ calculated as: $\left[- 4 + 2\right]$
The other at: $y = 6$ calculated as: $\left[+ 4 + 2\right]$

To summarise:

If we plot all of our points on the graph, we get:

Centre Point at: $\left(- 3 , 2\right)$

$x$ Intercepts at: $x = + \sqrt{21} - 3$ and $x = - \sqrt{21} - 3$
$y$ Intercepts at: $y = - 2$ and $y = 6$