Finding centre and radius of a circle (pre calc)

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jennielynn_1980

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Hi everyone! I am new here. I am taking a course in calculus for life sciences. Right now we are doing some preliminary stuff to get ready for the "real" calculus.
Here is the question that is giving me trouble:

Find the centre and radius of the circle given by the equation 0 = x[sup:1sfkrpth]2[/sup:1sfkrpth] + y[sup:1sfkrpth]2[/sup:1sfkrpth] - 4x + 2y -11
The answer is given as:
0 = (x[sup:1sfkrpth]2[/sup:1sfkrpth] -4x + 4) + (y[sup:1sfkrpth]2[/sup:1sfkrpth] + 2y +1) - 11 - 5
0 = (x[sup:1sfkrpth]2[/sup:1sfkrpth] -4x + 4) + (y[sup:1sfkrpth]2[/sup:1sfkrpth] + 2y +1) - 16
16 = ( x - 2)2 + ( y +1 )2
r = 4 (x[sub:1sfkrpth]0[/sub:1sfkrpth], y[sub:1sfkrpth]0[/sub:1sfkrpth]) = (2, -1)

What I am having trouble with is where the one and four come from in the first step. I understand the -5 is added at the end to account for the addition of the one and the four in the brackets but I am unsure where these numbers come from.
Thanks for your help :)
Jenn
 
jennielynn_1980 said:
Hi everyone! I am new here. I am taking a course in calculus for life sciences. Right now we are doing some preliminary stuff to get ready for the "real" calculus.
Here is the question that is giving me trouble:

Find the centre and radius of the circle given by the equation 0 = x[sup:2lvkt2gy]2[/sup:2lvkt2gy] + y[sup:2lvkt2gy]2[/sup:2lvkt2gy] - 4x + 2y -11
The answer is given as:
0 = (x[sup:2lvkt2gy]2[/sup:2lvkt2gy] -4x + 4) + (y[sup:2lvkt2gy]2[/sup:2lvkt2gy] + 2y +1) - 11 - 5
0 = (x[sup:2lvkt2gy]2[/sup:2lvkt2gy] -4x + 4) + (y[sup:2lvkt2gy]2[/sup:2lvkt2gy] + 2y +1) - 16
16 = ( x - 2)2 + ( y +1 )2
r = 4 (x[sub:2lvkt2gy]0[/sub:2lvkt2gy], y[sub:2lvkt2gy]0[/sub:2lvkt2gy]) = (2, -1)

What I am having trouble with is where the one and four come from in the first step. I understand the -5 is added at the end to account for the addition of the one and the four in the brackets but I am unsure where these numbers come from. <<< Have you done "completing the square" method - while solving quadratic equations?
Thanks for your help :)
Jenn
Click to expand...
 
Yes, if you are in a calc class then you are probably expected to know how to 'complete the square'.

Group the x's and y's together and leave room for the other term:

\(\displaystyle (x^{2}-4x+?)+(y^{2}+2y+?)=11\)

Now, look at the term in front of the x's. We have a -4 and a 2.

Take half that, square it, and add to both sides.

Half of -4 is -2, squared is 4. Half of 2 is 1, squared is 1.

We get \(\displaystyle (x^{2}-4x+4)+(y^{2}+2y+1)=11+4+1=16\)

Factor each one. They are perfect squares:

\(\displaystyle (x-2)^{2}+(y+1)^{2}=16\)

See there?. Not difficult?.

Now,what's the center and radius?.
 
The radius is 4 and the centre is (2, -1)

So just to be clear if the 2y had actually been 9y then the answer for that section would have been y[sup:22dfnywa]2[/sup:22dfnywa]+ 9y + 3??? or would it have to be something like y[sup:22dfnywa]2[/sup:22dfnywa] + 8y + 4?

And a big thanks to everyone for helping me out. I am little lost in this class because I have done any math for about 10 yrs :?
 
If the 2y would have been 9y, then half of 9 is 4.5, and 4.5 squared is 20.25. You'd have \(\displaystyle y^{2}+9y+20.25\)

When factored equals \(\displaystyle (y+\frac{9}{2})^{2}\)
 
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