Chapter 8.4, Problem 58E

Solution

To check: whether the given statement is true or false.

The given statement is true because the coefficients of $x^2$ and $y^2$ are same and there is no $xy$ -term in the given equation.

Given Information:

The given equation is $x^2+x^2+Dx+Ey+F=0$.

Calculation:

It is known that a circle equation has no $xy$ -term and the coefficients of $x^2$ and $y^2$ are the same.

It can be observed from the given equation that the coefficients of $x^2$ and $y^2$ are the same and there is no $xy$ -term in the given equation. Therefore, the given equation represents a circle equation.

Try to complete the square of $x$ and $y$ , and write the equation in the standard form of a circle equation.

  $\begin{array}{l}{x}^2+y^2+Dx+Ey+F=0\\ \left(x^2+Dx\right)+\left(y^2+Ey\right)=-F\\ \left(x^2+Dx+\frac{D^2}{4}\right)+\left(y^2+Ey+\frac{E^2}{4}\right)=-F+\frac{D^2}{4}+\frac{E^2}{4}\\ {\left(x+\frac{D}{2}\right)}^2+{\left(y+\frac{E}{2}\right)}^2=\frac{-4F+D^2+E^2}{4}\end{array}$

Thus, the given conic equation is a circle centered at $\left(\frac{-D}{2},\frac{E}{2}\right)$ with radius $\sqrt{\frac{D^2+E^2-4F}{4}}$ or a degenearte circle if $D^2+E^2-4F\le 0$ .

Therefore, the given statement is true.