# How do you graph  x^2-14x+y^2-14y+89=0?

Feb 2, 2016

Transform into the circle equation:

${\left(x - 7\right)}^{2} + {\left(y - 7\right)}^{2} = {3}^{2}$

and graph a circle with the center point $\left(7 , 7\right)$ and the radius $3$.

#### Explanation:

Your equation can be transformed as an equation of a circle.

The standard form of a circle equation is

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

where the center point of the circle is $\left({x}_{m} , {y}_{m}\right)$ and the radius of the circle is $r$.

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Let me show you how to transform your equation.

${x}^{2} - 14 x + {y}^{2} - 14 y + 89 = 0$

... subtract $89$ from both sides...

$\iff {x}^{2} - 14 x + {y}^{2} - 14 y = - 89$

Now, let's thrive for the term ${\left(x - {x}_{m}\right)}^{2}$ first.

Using the formula

${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$,

we already have two of the three necessary expressions: ${x}^{2} = {a}^{2}$, thus $x = a$ and $- 14 x = - 2 a b$, thus $b = 7$.

So, the missing term is ${b}^{2} = 49$. Let's add this term on both sides of the equation:

${x}^{2} - 14 x + {y}^{2} - 14 y = - 89$

$\iff {x}^{2} - 14 x \textcolor{red}{\text{ "+49) + y^2 - 14y = -89 color(red)(" } + 49}$

$\iff {\left(x - 7\right)}^{2} + {y}^{2} - 14 y = - 40$

Similarly, for the term ${\left(y - {y}_{m}\right)}^{2}$, there is $49$ missing as well. So, let's add $49$ to both sides of the equation again:

$\iff {\left(x - 7\right)}^{2} + {y}^{2} - 14 y \textcolor{b l u e}{\text{ "+49) = -40 color(blue)(" } + 49}$

$\iff {\left(x - 7\right)}^{2} + {\left(y - 7\right)}^{2} = 9$

$\iff {\left(x - 7\right)}^{2} + {\left(y - 7\right)}^{2} = {3}^{2}$

Now we have built a circle equation. The equation represents a circle with the center point $\left(7 , 7\right)$ and the radius $3$.

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Graph:

graph{(x-7)^2 + (y-7)^2 = 3^2 [-10.98, 21.06, -3.53, 12.49]}