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The problem you uploaded gives the general equation of a circle:

\[
x^2 + y^2 + 4x + 12y + 3 = 0
\]

It asks you to:
a) Determine the standard (or ordinary) equation of the circle,  
b) Find the value of \( h + c \), where \( h \) is the \( x \)-coordinate of the center, and \( c \) is the constant on the right-hand side of the standard form equation.

### Let's solve step by step.

#### Step 1: Rearrange into standard form
The standard form of a circle's equation is:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
Where \( (h, k) \) is the center of the circle and \( r \) is the radius.

To find this form, we will complete the square for the \( x \)- and \( y \)-terms in the given equation:

\[
x^2 + 4x + y^2 + 12y + 3 = 0
\]

##### Completing the square for \( x \):
\[
x^2 + 4x \quad \text{(Take half of 4, square it: \( \left(\frac{4}{2}\right)^2 = 4 \))}.
\]
Add and subtract 4:
\[
x^2 + 4x + 4 - 4 = (x + 2)^2 - 4
\]

##### Completing the square for \( y \):
\[
y^2 + 12y \quad \text{(Take half of 12, square it: \( \left(\frac{12}{2}\right)^2 = 36 \))}.
\]
Add and subtract 36:
\[
y^2 + 12y + 36 - 36 = (y + 6)^2 - 36
\]

Now substitute these completed square forms back into the original equation:
\[
(x + 2)^2 - 4 + (y + 6)^2 - 36 + 3 = 0
\]

Combine constants:
\[
(x + 2)^2 + (y + 6)^2 - 37 = 0
\]

Rearrange to get the standard form:
\[
(x + 2)^2 + (y + 6)^2 = 37
\]

This is the standard form of the equation. The center of the circle is \( (-2, -6) \), and the radius squared is \( 37 \).

#### Step 2: Find \( h + c \)
From the standard form equation \( (x + 2)^2 + (y + 6)^2 = 37 \), we know:
- \( h = -2 \) (since the center is \( (-2, -6) \))
- \( c = 37 \) (the constant on the right-hand side)

Therefore:
\[
h + c = -2 + 37 = 35
\]

#### Conclusion:
None of the options in the problem exactly match this result, but I can provide clarity on the calculations. If needed, I can recheck anything.

Let me know if you'd like to explore this further or have specific questions!

Here are 5 follow-up questions based on this solution:
1. What is the geometric significance of completing the square in circle equations?
2. How would the equation change if the radius was halved?
3. What is the general process for converting any conic section to standard form?
4. How do the signs of \( h \) and \( k \) in the standard form affect the location of the center?
5. Can a circle's equation ever have imaginary solutions for \( h \) or \( k \)?

**Tip**: Always check that the coefficients of \( x^2 \) and \( y^2 \) are 1 when completing the square for circles.

Determine the standard form of the circle equation: x^2 + y^2 + 4x + 12y + 3 = 0. Also, find the value of h + c where h is the x-coordinate of the center.

Learn how to convert the general equation of a circle, x^2 + y^2 + 4x + 12y + 3 = 0, to its standard form using the method of completing the square. The problem involves key geometry concepts and algebraic techniques, suitable for students in Grades 9-10.

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