Questions on Geometry: Circles and their properties answered by real tutors!

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Question 142614: The radius of a circle defined by x^2+y^2-2x-4y+1=0: The radius of a circle defined by x^2+y^2-2x-4y+1=0
Answer by solver91311(1877) About Me  (Show Source):
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The standard equation for a circle with radius r and center at (h,k) is (x-h)^2+(y-k)^2=r^2.

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^2-2x+y^2-4y=-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.

((-2)/2)^2=1 so x^2-2x+1+y^2-4y=-1+1

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

((-4)/2)^2=4 so x^2-2x+1+y^2-4y+4=-1+1+4

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

x^2-2x+1=(x-1)^2 and y^2-4y+4=(y-2)^2, so:

(x-1)^2+(y-2)^2=4

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

sqrt(4)=2

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