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In my exercise I was instructed to find the equation of the circle from the following:

$$4x^2+4y^2-24x-32y-4=0$$

In the solution, the $LHS$ is multiplied by $1/4$ for simplicity. Instead, I did the following:

$$4x^2-24x = (2x-6)^2 - 36$$

$$4y^2-32y = (2y-8)^2 - 64$$

Thus, I arrived at:

$$(2x-6)^2 + (2y-8)^2 = 104$$

$$((1/2)(x-3))^2+((1/2)(y-4))^2 = 104$$

$$1/4\ [\ (x-3)^2 + (y-4)^2\ ] = 104$$

Which leads me to the super wrong answer of:

$$(x-3)^2 + (y-4)^2 = 416$$

That circle is way too big, but my mistake here is unbeknownst to me. However, the center coordinates agree with the solution, while the radius is off by about $390$ or something.

Can someone point out the flaw in my algebra?

hardmath
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sangstar
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    You have taken out a factor $\frac 12$ instead of a factor $2$ and since these appear squared, you are out by a factor of $16$ – Mark Bennet Aug 12 '17 at 15:40
  • You factored a $2$ out of a square, where it should have been a $4$ in the step $((1/2)(x-3))^2+((1/2)(y-4))^2 = 104$ – sharding4 Aug 12 '17 at 15:41

4 Answers4

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dividing by $4$ we obtain $$x^2-6x+y^2-8y-1=0$$ compleeting to squares we get $$x^2-6x+9-9+y^2-8y+16-16=1$$ therefore we get $$(x-3)^2+(y-4)^2=26$$

Dr. Sonnhard Graubner
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Note that

\begin{align}(2x-6)^2 + (2y-8)^2 &= 104\iff (2(x-3))^2+(2(y-4))^2=104 \\&\iff 4(x-3)^2+4(y-4)^2=104 \\&\iff (x-3)^2+(y-4)^2=\frac{104}{4}=26.\end{align}

mfl
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Hint:

$$ (2x-6)^2=[2(x-3)]^2=4(x-3)^2 $$

Emilio Novati
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Hem, if you were asked to "find the equation of the circle", it is

$$4x^2+4y^2-24x-32y-4=0.$$