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

##### 1 Answer

#### Answer:

Transform into the circle equation:

#(x-7)^2 + (y-7)^2 = 3^2#

and graph a circle with the center point

#### Explanation:

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

The standard form of a circle equation is

#(x - x_m)^2 + (y - y_m)^2 = r^2# ,

where the center point of the circle is

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Let me show you how to transform your equation.

#x^2 - 14x + y^2 - 14y + 89 = 0#

... subtract

#<=> x^2 - 14x + y^2 - 14y = -89#

Now, let's thrive for the term

Using the formula

#(a - b)^2 = a^2 - 2ab + b^2# ,

we already have two of the three necessary expressions:

So, the missing term is

#x^2 - 14x + y^2 - 14y = -89#

#<=> x^2 - 14x color(red)(" "+49) + y^2 - 14y = -89 color(red)(" "+49)#

#<=> (x-7)^2 + y^2 - 14y = -40#

Similarly, for the term

#<=> (x-7)^2 + y^2 - 14y color(blue)(" "+49) = -40 color(blue)(" "+49)#

#<=> (x-7)^2 + (y-7)^2 = 9#

#<=> (x-7)^2 + (y-7)^2 = 3^2 #

Now we have built a circle equation. The equation represents a circle with the center point

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**Graph:**

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