Question

Find the center and radius of the circle defined by the equation x^2 + y^2 + 5x - 4y + 2 = 0.

Answer 1

You would have to complete the square on this in order to change it into the correct form. Any idea on how to complete the square?

Answer 2

I followed the steps from the textbook. The first step was to group the x and y terms and the constants to one side.
x^2 + 5x + __ + y^2 - 4y + __ = -2
Then to compete the square for both variables. This is where Im getting confused. They filled the first blank with 25/4 and the second blank with 4, i just dont know how they got that.

Answer 3

Well, they say half it and square it. Pretty easy mnemonic. Alright, so let's group these
(x^2 + 5x + __) + (y^2 - 4y + ___) = -2 + ___ + ___
Now we know quadratics come in the form of ax^2 + bx + c. Well, when I say half it and square it, this is the operation we do to the bx / by terms in this. So for us, the bx coefficient is 5 and the by coefficient is - 4. So these are what we half and square.
So half of 5 gives us 5/2. When we square that, we get 25/4. Now this number fills in the blank that is with the x portion. In order to keep the entire equation balanced, though, we must also add 25/4 to the other side of the equal sign. So far we would have this:
\[(x ^{2}+5x+\frac{ 25 }{ 4 }) +(y ^{2}-4y+---)=-2+\frac{ 25 }{ 4 }+---\]

Answer 4

So now we do the same thing with the y's. Our by coefficient is -4, so half it and square it. half of -4 would be -2, squared would be positive 4. So we add this 4 to both blanks remaining, one inside the y parenthesis and one on the other side of the equal sign so we can keep the equation balanced. This leaves us with:
\[(x ^{2}+5x+\frac{ 25 }{ 4 })+(y ^{2}-4y+4)=-2+\frac{ 25 }{ 4 }+4\]Now all we need to do is turn the parenthesis portions into the form of (x-h)^2 + (y-k)^2. To do this, we simply find what half of bx and by are. Have of bx is 5/2, so the parenthesis becomes this:
\[(x+\frac{ 5 }{ 2 })^{2}\]Half of the by term is -2, so now throwing it all together I have:
\[(x+\frac{ 5 }{ 2 })^{2}+(y-2)^{2}=-2+4+\frac{ 25 }{4 }\]
Adding the constants on the right. -2 + 4 = positive 2, which I can convert into 8/4 to make this a common denominator. Adding 8/4 with 25/4 leaves me with 33/4, so my final answer is:
\[(x+\frac{ 5 }{ 2 })^{2}+(y-2)^{2}=\frac{ 33 }{ 4 }\]

Now the center of a circle is found at (h,k). H and k are found in the formula in these positions:
(x-h)^2 + (y-k)^2 = r^2. Note how I put minus signs inside of the parenthesis. These MUST be minus signs. If you see a plus sign, it is because the number is actually negative. That being said, we have (x+5/2)^2. Since there is a plus sign, this means this is actually (x-(-5/2))^2. So the x-coordinate of our center is -5/2. As for the y part, it already is in y-k form, so the y coordinate of our center is just 2, meaning the center is at (-5/2, 2). Now the formula says that on the other side of the equal sing is r^2. this means if I want r, I need to take the square root of that side. This gives us;
\[\sqrt{\frac{ 33 }{ 4 }}=\frac{ \sqrt{33} }{ \sqrt{4} }=\frac{ \sqrt{33} }{ 2 }\]
So that should be all of the information needed for this problem

Answer 5

\[x^2+y^2+5x-4y+2=0\]
\[(x^2+5x+6.5)+(y^2-4y+4)=-2+4+6.5\]
\[(x+2.5)^2+(y-2)^2=8.5\]
C=(-2.5, 2) \[r=\sqrt{8.5}\]

Answer 6

Sorry, instead of 6.5 there should be 6.25 and the radius is \[\sqrt{8.25}\]