To find the center of the circle given by the equation x² + y² – 4x – 8y + 11 = 0, we need to rewrite the equation in standard form. The standard form of a circle’s equation is:
(x – h)² + (y – k)² = r²
where (h, k) is the center of the circle and r is the radius.
First, we rearrange the equation:
x² – 4x + y² – 8y + 11 = 0
Next, we move the constant to the other side:
x² – 4x + y² – 8y = -11
Now, we will complete the square for the x terms and the y terms.
For the x terms:
- Take half of -4 (which is -2), square it to get 4.
- Add and subtract 4 inside the equation.
For the y terms:
- Take half of -8 (which is -4), square it to get 16.
- Add and subtract 16 inside the equation.
So our equation now looks like this:
(x² – 4x + 4) + (y² – 8y + 16) = -11 + 4 + 16
This simplifies to:
(x – 2)² + (y – 4)² = 9
Now we see that the center (h, k) of the circle is (2, 4) and the radius squared, r², is 9. Therefore the radius r is 3.
Final Answer: The center of the circle is at the point (2, 4).