Center(3,-4) and passes through the origin

To find the equation of a circle with the center at \((3, -4)\) and passing through the origin \((0, 0)\), you can follow these steps:

### Step 1: Identify the center and radius of the circle.
- The center \((h, k)\) is given as \((3, -4)\).
- To find the radius \(r\), use the distance formula between the center and the point on the circle (the origin in this case).

\[
r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

Substitute the values:

\[
r = \sqrt{(0 - 3)^2 + (0 - (-4))^2} = \sqrt{(-3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\]

### Step 2: Write the equation of the circle.
The standard form of the equation of a circle is:

\[
(x - h)^2 + (y - k)^2 = r^2
\]

Substitute the values of \(h\), \(k\), and \(r\):

\[
(x - 3)^2 + (y + 4)^2 = 5^2
\]

Simplify:

\[
(x - 3)^2 + (y + 4)^2 = 25
\]

So, the equation of the circle is:

\[
(x - 3)^2 + (y + 4)^2 = 25
\]

Would you like more details on how this formula was derived or applied?

### Related Questions:
1. How do you derive the equation of a circle from general principles?
2. What is the geometric interpretation of the radius in this context?
3. How would the equation change if the circle passed through a different point?
4. How do you find the center and radius if only the equation is given?
5. How can you verify if a point lies inside, on, or outside a given circle?

### Tip:
When dealing with circles, remember that the distance from any point on the circle to its center is always constant, which is the radius. This property is key to many problems involving circles.

Learn how to find the equation of a circle with center (3, -4) that passes through the origin (0, 0). Step-by-step solution includes using the distance formula and the standard equation of a circle. Suitable for Grades 9-12.

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