How do you graph # x^2-14x+y^2-14y+89=0#?
1 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]}