Certainly! Let's go through the steps to convert the given equation of the circle from its general form to the standard form.
1. Given the general form of the circle's equation:
[tex]\[
7x^2 + 7y^2 - 28x + 42y - 35 = 0
\][/tex]
2. Simplify by dividing the entire equation by 7:
[tex]\[
x^2 + y^2 - 4x + 6y - 5 = 0
\][/tex]
3. Group the [tex]\( x \)[/tex] and [tex]\( y \)[/tex] terms for completing the square:
[tex]\[
(x^2 - 4x) + (y^2 + 6y) = 5
\][/tex]
4. Complete the square for the [tex]\( x \)[/tex]-terms and the [tex]\( y \)[/tex]-terms:
- For [tex]\( x \)[/tex]:
[tex]\[
x^2 - 4x \quad \text{can be rewritten as} \quad (x - 2)^2 - 4
\][/tex]
- For [tex]\( y \)[/tex]:
[tex]\[
y^2 + 6y \quad \text{can be rewritten as} \quad (y + 3)^2 - 9
\][/tex]
5. Rewrite the given equation with the completed squares:
[tex]\[
(x - 2)^2 - 4 + (y + 3)^2 - 9 = 5
\][/tex]
6. Combine constants on the right-hand side of the equation:
[tex]\[
(x - 2)^2 + (y + 3)^2 - 13 = 5
\][/tex]
[tex]\[
(x - 2)^2 + (y + 3)^2 = 18
\][/tex]
7. The equation in standard form is:
[tex]\[
(x - 2)^2 + (y + 3)^2 = 18
\][/tex]
8. Identify the center and radius of the circle:
- The center is [tex]\((2, -3)\)[/tex].
- The radius [tex]\( r \)[/tex] can be found from the equation [tex]\( r^2 = 18 \)[/tex]:
[tex]\[
r = \sqrt{18} \approx 4.242640687119285
\][/tex]
Now, if we go back to the multiple-choice selections:
- The standard form of the equation is [tex]\((x-2)^{\wedge} 2+(y+3)^{\wedge} 2=18\)[/tex]
- The center of the circle is at the point [tex]\((2, -3) \)[/tex]
- The radius is [tex]\(4.242640687119285\)[/tex] units.
So, the blank for the radius should be filled with [tex]\(4.242640687119285\)[/tex].
