1. What is the center of the circle of this equation $4x^2 + 4y^2 - 8x + 12y - 25 = 0$? 2. The length of the diameter of the circle represented by the equation $2x^2 + 2y^2 - 8 = 0... 1. What is the center of the circle of this equation $4x^2 + 4y^2 - 8x + 12y - 25 = 0$? 2. The length of the diameter of the circle represented by the equation $2x^2 + 2y^2 - 8 = 0$ is: Read more

Steps to Solve

1. Find the center of the circle First, we need to rewrite the equation $4x^2 + 4y^2 - 8x + 12y - 25 = 0$ in the standard form of a circle's equation, which is $(x-h)^2 + (y-k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius. Divide the entire equation by 4 to simplify: $x^2 + y^2 - 2x + 3y -\frac{25}{4} = 0$

2. Complete the square for x and y To complete the square for $x$, we take half of the coefficient of $x$ which is $-2$, so $\frac{-2}{2} = -1$ and square it to get $(-1)^2 = 1$. To complete the square for $y$, we take half of the coefficient of $y$ which is $3$, so $\frac{3}{2}$ and square it to get $(\frac{3}{2})^2 = \frac{9}{4}$. Add and subtract these values to the equation: $(x^2 - 2x + 1) + (y^2 + 3y + \frac{9}{4}) = \frac{25}{4} + 1 + \frac{9}{4}$

3. Rewrite the equation Rewrite the equation as: $(x-1)^2 + (y+\frac{3}{2})^2 = \frac{25}{4} + \frac{4}{4} + \frac{9}{4}$ $(x-1)^2 + (y+\frac{3}{2})^2 = \frac{38}{4} = \frac{19}{2}$

4. Identify the center of the circle The center of the circle is $(h, k) = (1, -\frac{3}{2}) = (1, -1.5)$. However, this answer is not among the possible options. Let's re-examine the completion of squares: $(x^2 - 2x + 1) + (y^2 + 3y + \frac{9}{4}) = \frac{25}{4} + 1 + \frac{9}{4}$ Combine terms on the right side: $\frac{25}{4} + \frac{4}{4} + \frac{9}{4} = \frac{38}{4} = \frac{19}{2}$ So $(x-1)^2 + (y+\frac{3}{2})^2 = \frac{19}{2}$ The center is $(1, -\frac{3}{2})$, which is $(1, -1.5)$. Still not among the options. Let us review the original equation: $4x^2 + 4y^2 - 8x + 12y - 25 = 0$ Divide by 4: $x^2 + y^2 - 2x + 3y - \frac{25}{4} = 0$ $(x^2 - 2x) + (y^2 + 3y) = \frac{25}{4}$ $(x^2 - 2x + 1) + (y^2 + 3y + \frac{9}{4}) = \frac{25}{4} + 1 + \frac{9}{4}$ $(x-1)^2 + (y + \frac{3}{2})^2 = \frac{25+4+9}{4} = \frac{38}{4} = \frac{19}{2}$ Therefore, the center is $(1, -\frac{3}{2})$. None of the options are correct

5. Find the length of the diameter of the circle The equation is $2x^2 + 2y^2 - 8 = 0$. Divide by 2: $x^2 + y^2 - 4 = 0$ $x^2 + y^2 = 4$ The radius is $r = \sqrt{4} = 2$. The diameter is $d = 2r = 2(2) = 4$.