# Question #bbc54

Apr 17, 2017

Type: Circle

The domain is:

$- 14 \le x \le 4$

The range is:

$- 13 \le y \le 5$

The circle will be symmetric about a line with any slope, m, of the form:

$y = m \left(x + 5\right) - 4$

The center is the point $\left(- 5 , - 4\right)$

#### Explanation:

From this reference on Conic Sections we obtain the General Cartesian Form:

$A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0 \text{ [1]}$

Please observe in the given equation:

${x}^{2} + {y}^{2} + 10 x + 8 y = 40 \text{ [2]}$

that $A = C \mathmr{and} B = 0$. The reference tells us that this identifies a circle.

The Cartesian form for the equation of a circle is:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2} \text{ [3]}$

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

I shall expand the squares of equation [3]:

${x}^{2} - 2 h x + {h}^{2} + {y}^{2} - 2 k y + {k}^{2} = {r}^{2} \text{ [4]}$

By matching the terms in equation [4] with equation [2], we observe that equation [2] does not have terms corresponding to ${h}^{2}$ and ${k}^{2}$, therefore, we shall add ${h}^{2} + {k}^{2}$ to both sides of the equation and arrange the terms on the left side in the same order as equation [4]:

${x}^{2} + 10 x + {h}^{2} + {y}^{2} + 8 y + {k}^{2} = 40 + {h}^{2} + {k}^{2} \text{ [5]}$

Matching terms in equation [4] with equation [5], we observe that the following must be true:

$- 2 h x = + 10 x$ and $- 2 k y = + 8 y$

Dividing the equation by $- 2 x \mathmr{and} - 2 y$, respectively, will allow us to discover the values of h and k:

$h = - 5$ and $k = - 4$

It follows that:

${h}^{2} = 25$ and ${k}^{2} = 16$

Substitute the for ${h}^{2} \mathmr{and} {k}^{2}$ values into equation [5]:

${x}^{2} + 10 x + 25 + {y}^{2} + 8 y + 16 = 40 + 25 + 16 \text{ [6]}$

We know that the left side of equation [6] will collapse into squares in the same form as equation [3] with $h = - 5 \mathmr{and} k = - 4$

${\left(x - - 5\right)}^{2} + {\left(y - - 4\right)}^{2} = 40 + 25 + 16 \text{ [7]}$

Please notice that you can write the $- -$ as plus $+$ but I discourage people from doing this, because it often causes errors, when identifying the center.

Simply the right side of equation [7] and write it as a square:

${\left(x - - 5\right)}^{2} + {\left(y - - 4\right)}^{2} = {9}^{2} \text{ [8]}$

The domain is:

$- 14 \le x \le 4$

The range is:

$- 13 \le y \le 5$

The circle will be symmetric about a line with any slope, m, of the form:

$y = m \left(x + 5\right) - 4$

The center is the point $\left(- 5 , - 4\right)$