https://file.aiquickdraw.com/mathbot/file/1723561988166osd7r354.jpg

Let's analyze and solve problem number 11 from the new image.

### Problem Statement
The problem is asking for the equation of a circle that is tangent to the x-axis and y-axis and passes through the point \((3, 4)\). The general form of the equation of the circle is \(Ax + By + C = 0\). The question asks for the sum of all possible values of \(C\).

### Steps to Solve

1. **Understand the Circle's Properties:**
   - Since the circle is tangent to both the x-axis and y-axis, its center lies on the line \(y = x\).
   - Let the center of the circle be \( (r, r) \) where \( r \) is the radius of the circle.

2. **Equation of the Circle:**
   - The general equation of the circle centered at \((r, r)\) with radius \(r\) is:
   \[
   (x - r)^2 + (y - r)^2 = r^2
   \]

3. **Substitute the Point (3,4):**
   - The circle passes through the point \( (3, 4) \), so substitute \(x = 3\) and \(y = 4\) into the equation:
   \[
   (3 - r)^2 + (4 - r)^2 = r^2
   \]

4. **Expand and Simplify:**
   - Expand both squares:
   \[
   (9 - 6r + r^2) + (16 - 8r + r^2) = r^2
   \]
   Simplify:
   \[
   25 - 14r + 2r^2 = r^2
   \]
   Subtract \(r^2\) from both sides:
   \[
   r^2 - 14r + 25 = 0
   \]

5. **Solve the Quadratic Equation:**
   - Use the quadratic formula to solve for \(r\):
   \[
   r = \frac{-(-14) \pm \sqrt{(-14)^2 - 4(1)(25)}}{2(1)}
   \]
   \[
   r = \frac{14 \pm \sqrt{196 - 100}}{2}
   \]
   \[
   r = \frac{14 \pm \sqrt{96}}{2}
   \]
   \[
   r = \frac{14 \pm 4\sqrt{6}}{2}
   \]
   \[
   r = 7 \pm 2\sqrt{6}
   \]

6. **Calculate the Possible Values of C:**
   - We have two possible values for \(r\): \(7 + 2\sqrt{6}\) and \(7 - 2\sqrt{6}\).
   - Since the circle is tangent to both axes, the equation in standard form for the circle is:
   \[
   x^2 + y^2 - 2xr - 2yr + r^2 = 0
   \]
   In the general linear form \(Ax + By + C = 0\), \(C = r^2\).

   So, calculate \(r^2\) for both values of \(r\):
   - For \(r = 7 + 2\sqrt{6}\):
   \[
   C_1 = (7 + 2\sqrt{6})^2 = 49 + 28\sqrt{6} + 24 = 73 + 28\sqrt{6}
   \]
   - For \(r = 7 - 2\sqrt{6}\):
   \[
   C_2 = (7 - 2\sqrt{6})^2 = 49 - 28\sqrt{6} + 24 = 73 - 28\sqrt{6}
   \]

   The sum of all possible values of \(C\) is:
   \[
   C_1 + C_2 = (73 + 28\sqrt{6}) + (73 - 28\sqrt{6}) = 146
   \]

So, the sum of all possible values of \(C\) is **146**.

Would you like to go over any specific step or have further questions?

### Related Questions:
1. How do you derive the formula for the distance from a point to a line?
2. What is the geometric significance of a circle tangent to both axes?
3. Can the center of the circle be located anywhere else for this problem?
4. How does the quadratic equation relate to finding the radius of the circle?
5. What are the steps to convert a quadratic equation to a circle's equation?
6. How does the discriminant of a quadratic equation help in finding real solutions?
7. What is the importance of completing the square in circle equations?
8. How would the problem change if the circle passed through a different point?

### Tip:
Always double-check the signs and coefficients in your quadratic equations to avoid errors when calculating roots or solving for variables.

Learn how to solve problem number 11 involving the equation of a circle tangent to the x-axis and y-axis, passing through point (3, 4). Explore step-by-step solutions, circle equation derivations, and geometric insights.

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