SOLUTION: The radius of a circle defined by x^2+y^2-2x-4y+1=0

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Question 142614: The radius of a circle defined by x^2+y^2-2x-4y+1=0
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
The standard equation for a circle with radius r and center at (h,k) is %28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2.

So, you have to complete the square for x and for y in your equation to put it into standard form. That way you can determine the radius directly.

Process:
Step 1: Rearrange the equation so that the constant is on the right and the variables are grouped on the left:

x%5E2-2x%2By%5E2-4y=-1

Step 2: Take the coefficient on the 1st degree x term, divide by 2 and square the result. Add that result to both sides of the equation.

%28%28-2%29%2F2%29%5E2=1 so x%5E2-2x%2B1%2By%5E2-4y=-1%2B1

Step 3: Repeat step 2 for the 1st degree y term.

%28%28-4%29%2F2%29%5E2=4 so x%5E2-2x%2B1%2By%5E2-4y%2B4=-1%2B1%2B4

Step 4: Factor the two trinomial parts of the left and collect terms on the right.

x%5E2-2x%2B1=%28x-1%29%5E2 and y%5E2-4y%2B4=%28y-2%29%5E2, so:

%28x-1%29%5E2%2B%28y-2%29%5E2=4

Step 5: Take the square root of the resulting constant term to find the radius

sqrt%284%29=2

Super-Double-Plus Extra Credit: Write the ordered pair that represents the center of this circle.