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What is the radius of a circle whose equation is [tex][tex]$x^2 + y^2 - 10x + 6y + 18 = 0$[/tex][/tex]?

A. 2 units
B. 4 units
C. 8 units
D. 16 units

Sagot :

To determine the radius of the circle given by the equation [tex]\( x^2 + y^2 - 10x + 6y + 18 = 0 \)[/tex], we need to convert this equation into 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 the radius.

### Step-by-Step Solution:

1. Group [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
[tex]\[ x^2 - 10x + y^2 + 6y + 18 = 0 \][/tex]

2. Complete the square for [tex]\(x\)[/tex]:
- The expression involving [tex]\(x\)[/tex] is [tex]\(x^2 - 10x\)[/tex].
- To complete the square, take half of the coefficient of [tex]\(x\)[/tex], square it, and add and subtract that value. Half of [tex]\(-10\)[/tex] is [tex]\(-5\)[/tex]. Squaring [tex]\(-5\)[/tex] gives [tex]\(25\)[/tex]. So we rewrite [tex]\(x^2 - 10x\)[/tex] as:
[tex]\[ x^2 - 10x + 25 - 25 \][/tex]
- This can be written as:
[tex]\[ (x - 5)^2 - 25 \][/tex]

3. Complete the square for [tex]\(y\)[/tex]:
- The expression involving [tex]\(y\)[/tex] is [tex]\(y^2 + 6y\)[/tex].
- To complete the square, take half of the coefficient of [tex]\(y\)[/tex], square it, and add and subtract that value. Half of [tex]\(6\)[/tex] is [tex]\(3\)[/tex]. Squaring [tex]\(3\)[/tex] gives [tex]\(9\)[/tex]. So we rewrite [tex]\(y^2 + 6y\)[/tex] as:
[tex]\[ y^2 + 6y + 9 - 9 \][/tex]
- This can be written as:
[tex]\[ (y + 3)^2 - 9 \][/tex]

4. Substitute back into the equation and simplify:
[tex]\[ (x - 5)^2 - 25 + (y + 3)^2 - 9 + 18 = 0 \][/tex]
- Combine constants on the right side of the equation:
[tex]\[ (x - 5)^2 + (y + 3)^2 - 25 - 9 + 18 = 0 \][/tex]
[tex]\[ (x - 5)^2 + (y + 3)^2 - 16 = 0 \][/tex]
[tex]\[ (x - 5)^2 + (y + 3)^2 = 16 \][/tex]

From the standard form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], we can identify that [tex]\(h = 5\)[/tex], [tex]\(k = -3\)[/tex] and [tex]\(r^2 = 16\)[/tex].

Therefore, [tex]\(r = \sqrt{16} = 4\)[/tex].

So, the radius of the circle is [tex]\( \boxed{4} \)[/tex] units.