SOLUTION: can someone please help me with this problem: Complete the square and write the equation in standard form. Then give the center and radius of the circle. x^2^+ y^2 - 2x - 4y

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: can someone please help me with this problem: Complete the square and write the equation in standard form. Then give the center and radius of the circle. x^2^+ y^2 - 2x - 4y       Log On


   

Question 159699: can someone please help me with this problem:
Complete the square and write the equation in standard form. Then give the center and radius of the circle.
x^2^+ y^2 - 2x - 4y - 4 = 0
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2%2B+y%5E2+-+2x+-+4y+-+4+=+0

First, put the constant term on the right:
x%5E2%2B+y%5E2+-+2x+-+4y+=+4

Divide the coefficient on the first degree x term by 2. 2%2F2+=+1.

Square the result and add that result to both sides. 1%5E2=1

x%5E2%2B+y%5E2+-+2x+-+4y+%2B+1+=+4+%2B+1

Repeat the process with the coefficient on the first degree y term

4%2F2=2, 2%5E2=4

x%5E2%2B+y%5E2+-+2x+-+4y+%2B+1+%2B+4+=+4+%2B+1+%2B+4

Rearrange the terms:

x%5E2-2x%2B1%2By%5E2-4y%2B4=9

Now factor the x terms:

x%5E2-2x%2B1=%28x-1%29%5E2

And factor the y terms:

y%5E2-4y%2B4=%28y-2%29%5E2

Put the pieces back together:

%28x-1%29%5E2%2B%28y-2%29%5E2=9

Now, use the fact that a circle of radius r centered at (h,k) has an equation %28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2 to determine that your circle is centered at (1,2) and has a radius of sqrt%289%29=3