MathRevolution wrote:
(x-a)^2 + (y-b)^2 = r^2 is the equation of a circle with radius r and center (a,b). What is the radius of the circle with equation x^2 + y^2 – 10x + 4y = 20?
A. 4
B. 5
C. 6
D. 7
E. 8
This is an interesting problem, because while I've seen 'completing the square' used in solutions for GMAT problems before, I'm not sure that I've seen it
required in an official problem. I'd be curious to see if anybody can show an official problem that requires you to do this.
Here's how it works: you need to take the equation you have, and put it in a form so that all of the variables are contained in the \((x-a)^2\) and the \((y-b)^2\) parts of the equation, and you only have numbers left over.
Focus on the x's first. The terms that have an x in them are \(x^2 - 10x\). You want to make that look like (x-something)^2.
Well, when you square (x-a), you end up with \(x-2ax+a^2\). So, 2a has to equal 10 in order for the expressions to look alike. That means we're looking at \((x-5)^2\).
\((x-5)^2\) equals\(x^2 - 10x + 25\). Therefore, \(x^2 - 10x\)
can be rewritten as \((x-5)^2-25\).
Similarly, for the y terms, we'll rewrite as \((y+2)^2\), which equals \(y^2 + 4y + 4\).
Therefore, \(y^2+4y\) can be rewritten as\((y+2)^2-4\).
Now, put those back into the original equation.
\(x^2 + y^2 – 10x + 4y = 20\)
\((x-5)^2 - 25 + (y+2)^2 - 4 = 20\)
\((x-5)^2 + (y+2)^2 = 20 + 4 + 25 = 49 = 7^2\)
So, the radius is 7.