Select the correct answer from each drop-down menu.

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][tex]$(2, -3)$[/tex][/tex], and its radius is [tex]\square[/tex] units.



Answer :

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].