Question

which of the following is not a true statement?
choose the incorrect statement
a. the equation, x² + 3y² - 6x + 6y + 7 = 0, represents a circle in general form.
b. a circle of the form x² + y² = 4 has a center at (0,0).
c. you can graph the equation of a circle if you know the radius and the center of the circle.
d. the equation, (x - 3)² + y² = 3, represents a circle in standard form.

Explanation:

Step1: Recall the general form of a circle

The general form of a circle is $x^{2}+y^{2}+Dx + Ey+F = 0$ where the coefficients of $x^{2}$ and $y^{2}$ are equal. In the equation $x^{2}+3y^{2}-6x + 6y+7 = 0$, the coefficients of $x^{2}$ and $y^{2}$ are 1 and 3 respectively, so it is not a circle.

Step2: Analyze option B

The standard - form of a circle is $(x - a)^{2}+(y - b)^{2}=r^{2}$, where $(a,b)$ is the center and $r$ is the radius. For $x^{2}+y^{2}=4$, which can be written as $(x - 0)^{2}+(y - 0)^{2}=2^{2}$, the center is $(0,0)$ and radius $r = 2$.

Step3: Analyze option C

If we know the center $(a,b)$ and radius $r$ of a circle (in standard form $(x - a)^{2}+(y - b)^{2}=r^{2}$), we can graph the circle. We can plot the center point and then use the radius to find points on the circle.

Step4: Analyze option D

The equation $(x - 3)^{2}+y^{2}=3$ is in the standard form $(x - a)^{2}+(y - b)^{2}=r^{2}$ with $a = 3$, $b = 0$ and $r=\sqrt{3}$.

Answer:

A. The equation, $x^{2}+3y^{2}-6x + 6y + 7 = 0$, represents a circle in general form.