To find the center and radius of the circle represented by the equation x² + 4x + y² + 6y = 4, we need to rewrite this equation in the standard form of a circle’s equation, which is:
(x – h)² + (y – k)² = r²
where (h, k) is the center of the circle and r is the radius.
First, we will rearrange the given equation:
x² + 4x + y² + 6y = 4
Next, we complete the square for the x terms and the y terms.
1. For the x terms: x² + 4x
To complete the square, we take half of the coefficient of x (which is 4), square it (2² = 4), and rewrite:
x² + 4x = (x + 2)² – 4
2. For the y terms: y² + 6y
Similarly, take half of the coefficient of y (which is 6), square it (3² = 9), and rewrite:
y² + 6y = (y + 3)² – 9
Putting it all together, we can rewrite the equation:
(x + 2)² – 4 + (y + 3)² – 9 = 4
(x + 2)² + (y + 3)² – 13 = 4
(x + 2)² + (y + 3)² = 17
Now, we have the equation in standard form:
(x – (-2))² + (y – (-3))² = 17
From this, we can identify the center and radius:
- Center: (-2, -3)
- Radius: √17
In conclusion, the center of the circle is (-2, -3) and the radius is √17.