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To find 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].
Here’s the step-by-step process to achieve that:
1. Start with the given equation:
[tex]\[ x^2 + y^2 - 10x + 6y + 18 = 0 \][/tex]
2. Rearrange the equation to group the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms together:
[tex]\[ x^2 - 10x + y^2 + 6y = -18 \][/tex]
3. Complete the square for the [tex]\(x\)[/tex] terms:
- Take the coefficient of [tex]\(x\)[/tex] (which is [tex]\(-10\)[/tex]), divide by 2 (to get [tex]\(-5\)[/tex]), and square it (to get [tex]\(25\)[/tex]).
- Rewrite [tex]\(x^2 - 10x\)[/tex] as [tex]\((x - 5)^2 - 25\)[/tex].
[tex]\[ x^2 - 10x = (x - 5)^2 - 25 \][/tex]
4. Complete the square for the [tex]\(y\)[/tex] terms:
- Take the coefficient of [tex]\(y\)[/tex] (which is [tex]\(6\)[/tex]), divide by 2 (to get [tex]\(3\)[/tex]), and square it (to get [tex]\(9\)[/tex]).
- Rewrite [tex]\(y^2 + 6y\)[/tex] as [tex]\((y + 3)^2 - 9\)[/tex].
[tex]\[ y^2 + 6y = (y + 3)^2 - 9 \][/tex]
5. Substitute these completed squares back into the equation:
[tex]\[ (x - 5)^2 - 25 + (y + 3)^2 - 9 = -18 \][/tex]
6. Combine the constants on the right-hand side:
[tex]\[ (x - 5)^2 + (y + 3)^2 - 34 = -18 \][/tex]
Add 34 to both sides to isolate the completed square terms:
[tex]\[ (x - 5)^2 + (y + 3)^2 = 16 \][/tex]
7. Compare this with the standard form of a circle’s equation [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex]:
- Here, [tex]\(r^2 = 16\)[/tex].
8. Solve for the radius [tex]\(r\)[/tex]:
[tex]\[ r = \sqrt{16} = 4 \][/tex]
Therefore, the radius of the circle is [tex]\(4\)[/tex] units. The correct answer is:
[tex]\[ \boxed{4} \][/tex]
Here’s the step-by-step process to achieve that:
1. Start with the given equation:
[tex]\[ x^2 + y^2 - 10x + 6y + 18 = 0 \][/tex]
2. Rearrange the equation to group the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms together:
[tex]\[ x^2 - 10x + y^2 + 6y = -18 \][/tex]
3. Complete the square for the [tex]\(x\)[/tex] terms:
- Take the coefficient of [tex]\(x\)[/tex] (which is [tex]\(-10\)[/tex]), divide by 2 (to get [tex]\(-5\)[/tex]), and square it (to get [tex]\(25\)[/tex]).
- Rewrite [tex]\(x^2 - 10x\)[/tex] as [tex]\((x - 5)^2 - 25\)[/tex].
[tex]\[ x^2 - 10x = (x - 5)^2 - 25 \][/tex]
4. Complete the square for the [tex]\(y\)[/tex] terms:
- Take the coefficient of [tex]\(y\)[/tex] (which is [tex]\(6\)[/tex]), divide by 2 (to get [tex]\(3\)[/tex]), and square it (to get [tex]\(9\)[/tex]).
- Rewrite [tex]\(y^2 + 6y\)[/tex] as [tex]\((y + 3)^2 - 9\)[/tex].
[tex]\[ y^2 + 6y = (y + 3)^2 - 9 \][/tex]
5. Substitute these completed squares back into the equation:
[tex]\[ (x - 5)^2 - 25 + (y + 3)^2 - 9 = -18 \][/tex]
6. Combine the constants on the right-hand side:
[tex]\[ (x - 5)^2 + (y + 3)^2 - 34 = -18 \][/tex]
Add 34 to both sides to isolate the completed square terms:
[tex]\[ (x - 5)^2 + (y + 3)^2 = 16 \][/tex]
7. Compare this with the standard form of a circle’s equation [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex]:
- Here, [tex]\(r^2 = 16\)[/tex].
8. Solve for the radius [tex]\(r\)[/tex]:
[tex]\[ r = \sqrt{16} = 4 \][/tex]
Therefore, the radius of the circle is [tex]\(4\)[/tex] units. The correct answer is:
[tex]\[ \boxed{4} \][/tex]
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