Question

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

Answer 1

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`.

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 `(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]}