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I've had trouble on this one problem for a couple days. Complete the square on the X and Y terms to find the center and radius of the circle. $x^2+2x+y^2-4y=-4\:\:$

LizHenson
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4 Answers4

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Hint: $(x^2+2x+1)+(y^2-4y+4)=-4+5$

E W H Lee
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$$ \begin{align*} x^2+2x+y^2-4y &= -4 \\ \Rightarrow x^2+2x+\left( \frac{2}{2} \right)^2+y^2-4y+\left( -\frac{4}{2} \right)^2 &=-4+\left( \frac{2}{2} \right)^2+\left( -\frac{4}{2} \right)^2 \\ \Rightarrow \,\,\,\,\, \,\left(x+\frac{2}{2}\right)^2\quad \,\,\, +\quad \,\,\,\,\left( y-\frac{4}{2} \right)^2\qquad &= -4+\left( \frac{2}{2} \right)^2+\left( -\frac{4}{2} \right)^2 \ldots \end{align*} $$

Here I have successfully completed the square. I will leave you to clean up the result. Two days is a long time to work with no help, best of luck.

J. W. Perry
  • 5,507
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What you need to do is make the terms on the LHS perfect squares and the radius on the RHS positive. You need to add a number on both sides accordingly.

How to find the number? The hint lies in the $x$ and $y$ terms. $2x=2 \cdot x\cdot1$ and $4y=2 \cdot y\cdot2$ So the number to add will be $1^2+2^2=1+4=5$ :)

Diya
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The equation of the circle with center at $(h,k)$ and radius $a$ is given by this form,

$$(x-h)^2+(y-k)^2=a^2$$

If you look at the your equation, you will see that it can be expressed in the above form, your equation is addition of two squares as shown below:

$(a+b)^2=\underbrace{(\text{first term})^2}+\underbrace{2\times(\text{first term})(\text{second term})}+\underbrace{(\text{second term})^2}$
we have$\hspace{40 pt}x^2 \hspace{80 pt}2.x.1 \hspace{80 pt}\text{missing}$
completing square is adding the $\color{red}{\text{square of half of coefficeint of the middle term}}$so that we can express an expression as a square of some expression

$(x+1)^2=x^2+2x+\color{red}1\Rightarrow x^2+2x=(x+1)^2-1...(I)$
$(y-2)^2=y^2-4y+\color{red}4\Rightarrow y^2-4y=(y-2)^2-4...(II)$

Add $(I)$ and $(II)$

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

$\therefore (x+1)^2-1+(y-2)^2-4=-4$

$\therefore(x+1)^2+(y-2)^2=1$,

$\therefore(x-(-1))^2+(y-2)^2=1$

compare this with the first equation of the circle and we get radius$=a=1, \text{center=} (h,k)=(-1,2)$

Vikram
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