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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 = 18[/tex].

The center of the circle is at the point [tex](2, -3)[/tex], and its radius is [tex]\boxed{3\sqrt{2}}[/tex] units.



Answer :

Sure, let's go through the steps to convert the general form of a circle's equation to its standard form, identify its center, and determine its radius.

### Step-by-Step Solution:

1. Start with the given general form of the equation:
[tex]\[ 7x^2 + 7y^2 - 28x + 42y - 35 = 0 \][/tex]

2. Simplify the equation by dividing everything by 7:
[tex]\[ x^2 + y^2 - 4x + 6y - 5 = 0 \][/tex]

3. Rewrite the equation to help with completing the square:
[tex]\[ x^2 - 4x + y^2 + 6y = 5 \][/tex]

4. Complete the square for the [tex]\(x\)[/tex]-terms:
[tex]\[ x^2 - 4x \][/tex]
To complete the square, take half of the coefficient of [tex]\(x\)[/tex] (which is [tex]\(-4\)[/tex]), square it [tex]\((\frac{-4}{2})^2 = 4\)[/tex], and add and subtract this value inside the equation:
[tex]\[ (x^2 - 4x + 4 - 4) = (x - 2)^2 - 4 \][/tex]

5. Complete the square for the [tex]\(y\)[/tex]-terms:
[tex]\[ y^2 + 6y \][/tex]
To complete the square, take half of the coefficient of [tex]\(y\)[/tex] (which is [tex]\(6\)[/tex]), square it [tex]\((\frac{6}{2})^2 = 9\)[/tex], and add and subtract this value inside the equation:
[tex]\[ (y^2 + 6y + 9 - 9) = (y + 3)^2 - 9 \][/tex]

6. Rewrite the equation with these completed squares:
[tex]\[ (x - 2)^2 - 4 + (y + 3)^2 - 9 = 5 \][/tex]

7. Combine constants and move them to the other side:
[tex]\[ (x - 2)^2 + (y + 3)^2 = 5 + 4 + 9 \][/tex]
[tex]\[ (x - 2)^2 + (y + 3)^2 = 18 \][/tex]

Thus, the standard form of the equation is:
[tex]\[ (x - 2)^2 + (y + 3)^2 = 18 \][/tex]

8. Identify the center of the circle [tex]\((h, k)\)[/tex]:
The center of the circle is located at [tex]\((2, -3)\)[/tex].

9. Determine the radius:
From the standard form equation [tex]\((x - 2)^2 + (y + 3)^2 = 18\)[/tex], the radius squared is [tex]\(18\)[/tex].
Therefore, the radius [tex]\(r\)[/tex] is:
[tex]\[ r = \sqrt{18} = 4.242640687119285 \][/tex]

### Final Answer:
- The standard form of the equation is [tex]\((x-2)^2 + (y+3)^2 = 18\)[/tex].
- The center of the circle is at the point [tex]\((2, -3)\)[/tex].
- The radius of the circle is approximately [tex]\(4.242640687119285\)[/tex] units.

This is how you convert the general form of a circle's equation to its standard form and determine the center and the radius.