Question
Which equation is the standard form of the equation of a circle drawn on the coordinate plane, derived from the pythagorean theorem
0,0
0,5
(-4,-3)
0,0
0,5
(-4,-3)
Answers
GPT-4o mini
The standard form of the equation of a circle in the coordinate plane is derived from the Pythagorean Theorem. The standard equation of a circle with center at \((h, k)\) and radius \(r\) is given by:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
Now, let's derive the equations using the given points:
1. **For the center at (0, 0)**:
- Here, \(h = 0\) and \(k = 0\). The equation simplifies to:
\[
x^2 + y^2 = r^2
\]
where \(r\) is the radius.
2. **For the center at (0, 5)**:
- Here, \(h = 0\) and \(k = 5\). The equation becomes:
\[
(x - 0)^2 + (y - 5)^2 = r^2
\]
which simplifies to:
\[
x^2 + (y - 5)^2 = r^2
\]
3. **For the center at (-4, -3)**:
- Here, \(h = -4\) and \(k = -3\). The equation of the circle is:
\[
(x + 4)^2 + (y + 3)^2 = r^2
\]
In summary, the standard form of the equation of a circle depends on the center (h, k) and the radius r, as shown in the equations above. Depending on the center chosen from your list, you would have the different equations provided.
\[
(x - h)^2 + (y - k)^2 = r^2
\]
Now, let's derive the equations using the given points:
1. **For the center at (0, 0)**:
- Here, \(h = 0\) and \(k = 0\). The equation simplifies to:
\[
x^2 + y^2 = r^2
\]
where \(r\) is the radius.
2. **For the center at (0, 5)**:
- Here, \(h = 0\) and \(k = 5\). The equation becomes:
\[
(x - 0)^2 + (y - 5)^2 = r^2
\]
which simplifies to:
\[
x^2 + (y - 5)^2 = r^2
\]
3. **For the center at (-4, -3)**:
- Here, \(h = -4\) and \(k = -3\). The equation of the circle is:
\[
(x + 4)^2 + (y + 3)^2 = r^2
\]
In summary, the standard form of the equation of a circle depends on the center (h, k) and the radius r, as shown in the equations above. Depending on the center chosen from your list, you would have the different equations provided.