The graph of [tex][tex]$4x^2 + 4y^2 - 24x - 32y + 72 = 0$[/tex][/tex] is a circle. What is the radius of the circle?

A. [tex]\sqrt{7}[/tex]
B. 5
C. [tex]2\sqrt{7}[/tex]
D. 7



Answer :

To find the radius of the circle described by the equation [tex]\(4x^2 + 4y^2 - 24x - 32y + 72 = 0\)[/tex], we need to convert this equation into the standard form of a circle, [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex].

Here is the step-by-step process:

1. Divide the entire equation by 4 to simplify it:
[tex]\[ x^2 + y^2 - 6x - 8y + 18 = 0 \][/tex]

2. Rewrite the equation to group the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
[tex]\[ (x^2 - 6x) + (y^2 - 8y) = -18 \][/tex]

3. Complete the square for the [tex]\(x\)[/tex] terms:
- Take the coefficient of [tex]\(x\)[/tex], which is [tex]\(-6\)[/tex], halve it to get [tex]\(-3\)[/tex], and then square it to get 9.
- Add and subtract this square inside the equation:
[tex]\[ (x^2 - 6x + 9 - 9) + (y^2 - 8y) = -18 \][/tex]
[tex]\[ (x - 3)^2 - 9 + (y^2 - 8y) = -18 \][/tex]

4. Complete the square for the [tex]\(y\)[/tex] terms:
- Take the coefficient of [tex]\(y\)[/tex], which is [tex]\(-8\)[/tex], halve it to get [tex]\(-4\)[/tex], and then square it to get 16.
- Add and subtract this square inside the equation:
[tex]\[ (x - 3)^2 - 9 + (y^2 - 8y + 16 - 16) = -18 \][/tex]
[tex]\[ (x - 3)^2 - 9 + (y - 4)^2 - 16 = -18 \][/tex]

5. Combine the constants on the right-hand side of the equation:
[tex]\[ (x - 3)^2 + (y - 4)^2 - 25 = -18 \][/tex]
[tex]\[ (x - 3)^2 + (y - 4)^2 = 7 \][/tex]

The resulting equation [tex]\((x - 3)^2 + (y - 4)^2 = 7\)[/tex] is now in the standard form of a circle with the center [tex]\((h, k) = (3, 4)\)[/tex] and radius [tex]\(r\)[/tex].

From [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], we see that [tex]\(r^2 = 7\)[/tex].

Thus, the radius [tex]\(r\)[/tex] is:
[tex]\[ r = \sqrt{7} \][/tex]

Given the options:
- [tex]\(\sqrt{7}\)[/tex]
- 5
- [tex]\(2 \sqrt{7}\)[/tex]
- 7

The correct answer is [tex]\(\sqrt{7}\)[/tex].