# Determine the centre and radius of a circle

• Sep 10th 2011, 02:40 AM
Lepzed
Determine the centre and radius of a circle
Given is the following equation of a circle, for which I've to determine its centre and radius. I'm running into some difficulties when I'm trying to complete the squares.

$\displaystyle x^2 + y^2 -2x + 4y = 4 \Leftrightarrow (x - 1)^2 + (y + 2)^2 = 4$
This leads me to believe the centre would be (1, -2). Now to find its radius I'll just complete the squares and simplify:
$\displaystyle (x - 1)^2 + (y + 2)^2 = 4 \Leftrightarrow x^2 - 2x + 1 + y^2 + 4y + 4 = 4 \Leftrightarrow x^2 - 2x + y^2 + 4y = -1$

This obviously isn't correct, -1 being kind of a weird radius (Giggle). So what am I doing wrong here? The correct answer (according to my textbook) should be r=3. I guess I'm missing something with completing the y-part, since -1 + 4 = 3, but can't wrap my head around it...

Any handy tips for solving these kind of exercises?
• Sep 10th 2011, 02:47 AM
e^(i*pi)
Re: Determine the centre and radius of a circle
Quote:

Originally Posted by Lepzed
Given is the following equation of a circle, for which I've to determine its centre and radius. I'm running into some difficulties when I'm trying to complete the squares.

$\displaystyle x^2 + y^2 -2x + 4y = 4 \Leftrightarrow (x - 1)^2 + (y + 2)^2 = 4$
This leads me to believe the centre would be (1, -2). Now to find its radius I'll just complete the squares and simplify:
$\displaystyle (x - 1)^2 + (y + 2)^2 = 4 \Leftrightarrow x^2 - 2x + 1 + y^2 + 4y + 4 = 4 \Leftrightarrow x^2 - 2x + y^2 + 4y = -1$

This obviously isn't correct, -1 being kind of a weird radius (Giggle). So what am I doing wrong here? The correct answer (according to my textbook) should be r=3. I guess I'm missing something with completing the y-part, since -1 + 4 = 3, but can't wrap my head around it...

Any handy tips for solving these kind of exercises?

Your problem is that you're saying that $\displaystyle (x-1)^2 = x^2-2x$ when it does not. The answer is $\displaystyle x^2-2x = x^2 - 2x +1 - 1 = (x-1)^2 - 1$

In other words once you've completed the square on the LHS you should have $\displaystyle (x-1)^2 -1 + (y+2)^2 - 4 = 4$ and you can then add 5 to both sides to remove the constants: $\displaystyle (x-1)^2 + (y+2)^2 = 9$ which gives the radius as 3.

Your co-ordinates for the centre of the circle are correct
• Sep 10th 2011, 02:52 AM
Lepzed
Re: Determine the centre and radius of a circle
Ah, I see! Thanks!
• Sep 10th 2011, 04:47 AM
Lepzed
Re: Determine the centre and radius of a circle
I'm practicing some more and stumbled upon the following: describe the region defined by the following pair of inequalities:

$\displaystyle x^2 + y^2 < 2x, x^2 + y^2 < 2y$
I'm having some difficulties interpreting this. At first I thought I could rewrite each inequality to try to make some more sense of it, like so: $\displaystyle x^2 + y^2 < 2x \Leftrightarrow x^2 + y^2 - 2x < 0$, but that still doesn't make alot of sense to me.

What I've further tried, is to just write it as a 'normal' equation: $\displaystyle x^2 + y^2 = 2x$, then x & y are $\displaystyle \sqrt2$, same for $\displaystyle X^2 + y^2 = 2y$. Don't these circles describe exactly the same area?
If not, how should I deal with the 2x and 2y?
• Sep 10th 2011, 04:57 AM
Plato
Re: Determine the centre and radius of a circle
Quote:

Originally Posted by Lepzed
$\displaystyle x^2 + y^2 < 2x, x^2 + y^2 < 2y$
I'm having some difficulties interpreting this. At first I thought I could rewrite each inequality to try to make some more sense of it, like so: $\displaystyle x^2 + y^2 < 2x \Leftrightarrow x^2 + y^2 - 2x < 0$, but that still doesn't make alot of sense to me.

That system can be rewritten by completing squares.
$\displaystyle \left\{ \begin{gathered} (x - 1)^2 + y^2 < 1 \hfill \\ x^2 + (y - 1)^2 < 1 \hfill \\ \end{gathered} \right.$

Those are both the interiors of circles, open disks.