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To write the equation [tex]\( x^2 + y^2 - 18x + 8y + 5 = 0 \)[/tex] in standard form, we follow a step-by-step process to transform it into the form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex].
1. Group terms and move the constant to the other side of the equation:
[tex]\[ x^2 - 18x + y^2 + 8y = -5 \][/tex]
2. Determine the values that need to be added to both sides of the equation to complete the square for both the [tex]\(x\)[/tex]-terms and [tex]\(y\)[/tex]-terms:
[tex]\[ \left(-\frac{18}{2}\right)^2 = 81 \quad \text{and} \quad \left(\frac{8}{2}\right)^2 = 16 \][/tex]
3. Add these values to both sides of the equation:
[tex]\[ x^2 - 18x + 81 + y^2 + 8y + 16 = -5 + 81 + 16 \][/tex]
Simplifying the right side, we get:
[tex]\[ x^2 - 18x + 81 + y^2 + 8y + 16 = 92 \][/tex]
4. Write each trinomial as a binomial squared:
[tex]\[ (x - 9)^2 + (y + 4)^2 = 92 \][/tex]
Thus, the standard form of the equation of the circle given by [tex]\( x^2 + y^2 - 18x + 8y + 5 = 0 \)[/tex] is:
[tex]\[ (x - 9)^2 + (y + 4)^2 = 92 \][/tex]
1. Group terms and move the constant to the other side of the equation:
[tex]\[ x^2 - 18x + y^2 + 8y = -5 \][/tex]
2. Determine the values that need to be added to both sides of the equation to complete the square for both the [tex]\(x\)[/tex]-terms and [tex]\(y\)[/tex]-terms:
[tex]\[ \left(-\frac{18}{2}\right)^2 = 81 \quad \text{and} \quad \left(\frac{8}{2}\right)^2 = 16 \][/tex]
3. Add these values to both sides of the equation:
[tex]\[ x^2 - 18x + 81 + y^2 + 8y + 16 = -5 + 81 + 16 \][/tex]
Simplifying the right side, we get:
[tex]\[ x^2 - 18x + 81 + y^2 + 8y + 16 = 92 \][/tex]
4. Write each trinomial as a binomial squared:
[tex]\[ (x - 9)^2 + (y + 4)^2 = 92 \][/tex]
Thus, the standard form of the equation of the circle given by [tex]\( x^2 + y^2 - 18x + 8y + 5 = 0 \)[/tex] is:
[tex]\[ (x - 9)^2 + (y + 4)^2 = 92 \][/tex]
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