Question

Calculate the center of the circle by completing the square of the equation x2+y2−6x−12y+5=0
𝑥
2
+
𝑦
2
−
6
𝑥
−
12
𝑦
+
5
=
0
.

Answer 1

To find the center of the circle given by the equation \( x^2 + y^2 - 6x - 12y + 5 = 0 \), we will complete the square for the \( x \) and \( y \) terms.

1. **Rearranging the equation**:
Start by moving the constant term to the other side:
\[
x^2 + y^2 - 6x - 12y = -5
\]

2. **Completing the square for \( x \)**:
Take the \( x \) terms \( x^2 - 6x \):
- Half of \( -6 \) is \( -3 \).
- Squaring \( -3 \) gives \( 9 \).
- Thus, we can rewrite \( x^2 - 6x \) as:
\[
(x - 3)^2 - 9
\]

3. **Completing the square for \( y \)**:
Take the \( y \) terms \( y^2 - 12y \):
- Half of \( -12 \) is \( -6 \).
- Squaring \( -6 \) gives \( 36 \).
- Thus, we can rewrite \( y^2 - 12y \) as:
\[
(y - 6)^2 - 36
\]

4. **Substituting back into the equation**:
Replace \( x^2 - 6x \) and \( y^2 - 12y \) in the rearranged equation:
\[
(x - 3)^2 - 9 + (y - 6)^2 - 36 = -5
\]

Combine the constants:
\[
(x - 3)^2 + (y - 6)^2 - 45 = -5
\]
\[
(x - 3)^2 + (y - 6)^2 = 40
\]

5. **Identifying the center and radius**:
The equation now has the standard form of a circle \( (x - h)^2 + (y - k)^2 = r^2 \), where \( (h, k) \) is the center and \( r \) is the radius.
- Here, \( h = 3 \) and \( k = 6 \).

Thus, the center of the circle is at:
\[
\boxed{(3, 6)}
\]