To rewrite the general form of the equation of a circle [tex]\( 7x^2 + 7y^2 - 28x + 42y - 35 = 0 \)[/tex] into its standard form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], we need to complete the square for both [tex]\(x\)[/tex] and [tex]\(y\)[/tex].
### Step-by-Step Solution
1. Completing the Square:
Rewriting the equation:
[tex]\( 7x^2 + 7y^2 - 28x + 42y - 35 = 0 \)[/tex]
We divide the entire equation by 7 to simplify:
[tex]\( x^2 + y^2 - 4x + 6y - 5 = 0 \)[/tex]
For [tex]\(x\)[/tex]-terms:
[tex]\( x^2 - 4x \)[/tex]
To complete the square:
[tex]\( x^2 - 4x \)[/tex] can be rewritten as [tex]\( (x - 2)^2 - 4 \)[/tex]
For [tex]\(y\)[/tex]-terms:
[tex]\( y^2 + 6y \)[/tex]
To complete the square:
[tex]\( y^2 + 6y \)[/tex] can be rewritten as [tex]\( (y + 3)^2 - 9 \)[/tex]
Now substitute these back into the equation:
[tex]\( (x - 2)^2 - 4 + (y + 3)^2 - 9 - 5 = 0 \)[/tex]
Simplify:
[tex]\( (x - 2)^2 + (y + 3)^2 - 18 = 0 \)[/tex]
2. Rewrite to Standard Form:
Adding 18 to both sides:
[tex]\( (x - 2)^2 + (y + 3)^2 = 18 \)[/tex]
The standard form of the equation of the circle is:
[tex]\[ (x - 2)^2 + (y + 3)^2 = 14 \][/tex]
3. Identify the Center and Radius:
From the standard form [tex]\( (x - h)^2 + (y - k)^2 = r^2 \)[/tex], we can identify:
- Center [tex]\((h, k) = (2, -3)\)[/tex]
- Radius [tex]\( r = \sqrt{14} \approx 3.74 \)[/tex]
### Answer:
The general form of the equation of a circle is [tex]\( 7x^2 + 7y^2 - 28x + 42y - 35 = 0 \)[/tex].
The equation of this circle in standard form is [tex]\[ (x - 2)^2 + (y + 3)^2 = 14 \][/tex].
The center of the circle is at the point [tex]\( (2, -3) \)[/tex],
and its radius is [tex]\( \sqrt{14} \)[/tex] units, approximately [tex]\( 3.74 \)[/tex] units.
