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What is the radius of a circle whose equation is [tex]\(x^2 + y^2 + 8x - 6y + 21 = 0\)[/tex]?

A. 2 units
B. 3 units
C. 4 units
D. 5 units

Sagot :

To determine the radius of the circle given by the equation [tex]\( x^2 + y^2 + 8x - 6y + 21 = 0 \)[/tex], we need to rewrite the equation in the standard form of a circle's equation, which is [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], where [tex]\((h, k)\)[/tex] is the center of the circle and [tex]\(r\)[/tex] is its radius.

Here's the step-by-step process:

1. Rewrite the equation:
Start with the given equation:
[tex]\[ x^2 + y^2 + 8x - 6y + 21 = 0 \][/tex]
Move the constant term to the right side of the equation:
[tex]\[ x^2 + y^2 + 8x - 6y = -21 \][/tex]

2. Complete the square for the [tex]\(x\)[/tex] terms:
- Take the [tex]\(x\)[/tex] terms: [tex]\( x^2 + 8x \)[/tex]
- To complete the square, add and subtract [tex]\((\frac{8}{2})^2\)[/tex] which is [tex]\(16\)[/tex]:
[tex]\[ x^2 + 8x = (x + 4)^2 - 16 \][/tex]

3. Complete the square for the [tex]\(y\)[/tex] terms:
- Take the [tex]\(y\)[/tex] terms: [tex]\( y^2 - 6y \)[/tex]
- To complete the square, add and subtract [tex]\((\frac{-6}{2})^2\)[/tex] which is [tex]\(9\)[/tex]:
[tex]\[ y^2 - 6y = (y - 3)^2 - 9 \][/tex]

4. Substitute back into the equation:
Substitute the completed squares back into the equation:
[tex]\[ (x + 4)^2 - 16 + (y - 3)^2 - 9 = -21 \][/tex]

5. Simplify:
Combine like terms and move constants to the right side:
[tex]\[ (x + 4)^2 + (y - 3)^2 - 25 = -21 \][/tex]
[tex]\[ (x + 4)^2 + (y - 3)^2 = 4 \][/tex]

6. Identify the radius:
The equation [tex]\((x + 4)^2 + (y - 3)^2 = 4\)[/tex] is now in the standard form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex]. Here, [tex]\( (h, k) = (-4, 3) \)[/tex] is the center, and [tex]\( r^2 = 4 \)[/tex]. The radius [tex]\( r \)[/tex] is the square root of 4:
[tex]\[ r = \sqrt{4} = 2 \][/tex]

Therefore, the radius of the circle is [tex]\( 2 \)[/tex] units.