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

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Answer 1 · 2016-02-02T17:50:52

Transform into the circle equation:

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

and graph a circle with the center point #(7 , 7)# and the radius #3#.

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 #(x_m, y_m)# and the radius of the circle is #r#.

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Let me show you how to transform your equation.

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

... subtract #89# from both sides...

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

Now, let's thrive for the term #(x-x_m)^2# first.

Using the formula

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

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

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

#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 #(y - y_m)^2#, there is #49# missing as well. So, let's add #49# to both sides of the equation again:

#<=> (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 #(7, 7)# 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]}