Determine if $-2x-4x^2+12y = 4y^2 -1$ represents a circle

Question

I was given the following question on a practice test:

Determine whether the following equation represents a circle. If so, find its centre and radius. If not, list all reasons why. $$-2x-4x^2+12y = 4y^2 -1$$

The instructor posted the solution

$$h=-\frac14 \qquad k=\frac32 \qquad r=\sqrt{\frac{41}{16}}=\frac{\sqrt{41}}{4}$$

But didn't give details on how to solve it.

It would be great if someone could explain to me the steps of the solution, as I'm having trouble getting it into the form of a circle equation.

Answer 1

Rearranging some things:

$4x^2+2x + 4y^2-12y = 1$

Dividing by $4$ and adding $\square$'s as placeholders for values that we want to see in the future to "complete the square" so that we can more conveniently represent things.

$x^2+\frac{1}{2}x+\color{blue}\square + y^2-3y+\color{red}\square = \frac{1}{4}+\color{blue}\square+\color{red}\square$

Now, we want to pick a value for $\color{blue}\square$ such that the first three terms wind up in the form $(x+a)^2=x^2+2ax+a^2$.

Well, if since the $x$ term that we do already have is $\frac{1}{2}$ and since we want it to look like $2ax$, by setting them equal $2ax = \frac{1}{2}x$, and then dividing both sides by $2x$ we get $a=\frac{1}{4}$, so we will use a value of $a^2=(\frac{1}{4})^2=\frac{1}{16}$ in place of our $\color{blue}\square$. In doing so, we have:

$x^2+2\cdot \frac{1}{4}x + (\frac{1}{4})^2 + y^2-3y+\color{red}\square=\frac{1}{4}+(\frac{1}{4})^2+\color{red}\square$

Factoring the quadratic in $x$

$(x+\frac{1}{4})^2 + y^2-3y+\color{red}\square = \frac{1}{4}+(\frac{1}{4})^2+\color{red}\square$

And writing the quadratic in $x$ with a minus instead so that it is in the desired form

$(x-(-\frac{1}{4}))^2 + y^2-3y+\color{red}\square = \frac{1}{4}+(\frac{1}{4})^2+\color{red}\square$

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I leave completing the square for $y$ to you now that you have seen how it was performed for $x$.

You will wind up learning that this is going to be a circle, and it will be centered around the point $(-\frac{1}{4},y_0)$ with radius $\sqrt{\frac{1}{4}+(\frac{1}{4})^2+\color{red}\square}$ where I leave calculating $y_0$ and $\color{red}\square$ to you to finish.

Answer 2

You can also use unknown coefficients method as follows:

Suppose equation of circle is:

$(x-a)^2+(y-b)^2=r^2$

$x^2-2ax+a^2+y^2-2by+b^2=r^2$

Now we rewrite equation and compare the coefficient of powers of x and y and constant of two equation :

$x^2+\frac{1}{2}x+y^2-3y=\frac{1}{4}$

So we must have:

$-2a=\frac{1}{2}$ ⇒ $a=-\frac{1}{4}$

$-2b=-3$ ⇒ $b=\frac{3}{2}$

$r^2-a^2-b^2=\frac{1}{4}$ ⇒$r^2=(-\frac{1}{4})^2+(\frac{3}{2})^2+\frac{1}{4}=\frac{41}{16}$

⇒ $r=\frac{\sqrt{41}}{4}$