Need to prove an equation's plot is a circle, and say where it will be centered and what the radius is.

Question

The equation is the following: $u = \alpha \mu -\beta \sigma^2 -\beta \mu^2$ Where $\alpha>0$ and $\beta>0$ The u is fixed (drawing indifference curves). I need to prove the resulting plot is a circle and find the centre and radius of said circle. $\mu$ is on the y axis and $\sigma$ on the x axis. Thank you!

Answer 1

Recall that the formula for a circle is $(x-h)^2+(y-k)^2 = c^2$ where $(h,k)$ is the center of the circle and $c$ is the radius. With this in mind, the first thing we do is divide by $-\beta$ to get,

$$\frac{u}{-\beta}=\frac{\alpha}{-\beta}\mu+\mu^2 +\sigma^2$$

Now we need to complete the square for our polynomial in $\mu$:

$$\frac{u}{-\beta}=\left(\frac{\alpha}{-2\beta}\right)^2-\left(\frac{\alpha}{-2\beta}\right)^2+ \frac{\alpha}{-\beta}\mu+\mu^2 +\sigma^2$$

Now we can rewrite the polynomial in $\mu$ as a square:

$$\frac{u}{-\beta}=-\left(\frac{\alpha}{-2\beta}\right)^2+\left(\mu+\frac{\alpha}{-2\beta}\right)^2 +\sigma^2$$

and subtract the constant term to the left to get the form of the circle:

$$\frac{u}{-\beta}+\left(\frac{\alpha}{-2\beta}\right)^2=\left(\mu-\frac{\alpha}{2\beta}\right)^2 +\sigma^2$$

Now you can determine where the center is and what the radius is using the equation of a circle given at the beginning.

Answer 2

Rewrite the equation as $\beta\sigma^2+\beta\mu^2-\alpha\mu-u=0$. This fits the template of a general conic equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$. Its discriminant is equal to $-4\beta^2$, which is less than zero, so this equation represents an ellipse. The coefficients of $\sigma^2$ and $\mu^2$ are equal, so the ellipse has equal axis lengths: it’s a circle.

Its center can be found by inspection, but another way to find it is to set the gradient of the left-hand side to zero and solve for $\sigma$ and $\mu$. This leads to the system of equations $$2\beta\sigma = 0 \\ 2\beta\mu-\alpha = 0$$ with solution $\sigma = 0$, $\mu = {\alpha\over2\beta}$. This method works for any conic with a center—an ellipse or hyperbola. Applying this method to the general equation yields the center $\left({2CD-BE\over B^2-4AC},{2AE-BD\over B^2-4AC}\right)$.

To find the radius, you can convert the equation into the form $\sigma^2+\left(\mu-\frac\alpha{2\beta}\right)^2-r^2=0$ by dividing through by $\beta$ and completing the square, but there’s a short-cut you can use. It turns out that if you translate the equation by substituting $x'+h$ for $x$ and $y'+k$ for $y$ in the general equation of a conic, the resulting constant term $F'$ is equal to the value that you get by plugging $h$ and $k$ into the left-hand side of the equation. If $(h,k)$ are the coordinates of the conic’s center, this will eliminate the linear terms, and for a circle you’ll be left with an equation of the form $A(x'^2+y'^2)+F'=0$. The radius of this circle is then clearly $(-F'/A)^{\frac12}$. Applying this to the equation at top produces $\left(\frac u\beta+{\alpha^2\over4\beta^2}\right)^{\frac12}$ for the radius.