Question - Converting Circle Equation to Center-Radius Form
Solution:
The equation written in the image appears to be:\[ x^2 + y^2 + 2x - 4y - 11 = 0 \]This looks to be the equation of a circle in the general form, which can be converted into the center-radius form by completing the square for both x and y.The general equation for a circle is:\[ (x - h)^2 + (y - k)^2 = r^2 \]Where (h,k) is the center of the circle and r is its radius.Let's complete the square for the given equation:1. Group x and y terms together and add and subtract the necessary constants inside the parenthesis to form a perfect square trinomial:\[ (x^2 + 2x + 1) + (y^2 - 4y + 4) - 1 - 4 - 11 = 0 \]2. Complete the square by adding and subtracting the right constants inside each parenthesis:\[ (x^2 + 2x + 1) + (y^2 - 4y + 4) = 1 + 4 + 11 \]3. Rewrite as squares and simplify the right side of the equation:\[ (x + 1)^2 + (y - 2)^2 = 16 \]Thus, the center-radius form of the given circle's equation is:\[ (x + 1)^2 + (y - 2)^2 = 4^2 \]So the center of the circle is (-1, 2) and the radius is 4.
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