Question
what is the rate and slope intercept form of x^2 + y^2 - 16x - 4y + 32 = 0
Answers
GPT-4o mini
To find the slope-intercept form of the equation \(x^2 + y^2 - 16x - 4y + 32 = 0\), we first need to rewrite the equation in terms of \(y\).
1. Rearranging the equation:
\[
y^2 - 4y + x^2 - 16x + 32 = 0
\]
2. We will complete the square for both \(x\) and \(y\).
**Completing the square for \(x\)**:
- The expression \(x^2 - 16x\) can be completed as follows:
\[
x^2 - 16x = (x - 8)^2 - 64
\]
**Completing the square for \(y\)**:
- The expression \(y^2 - 4y\) can be completed as follows:
\[
y^2 - 4y = (y - 2)^2 - 4
\]
3. Now substituting back into the equation:
\[
(x - 8)^2 - 64 + (y - 2)^2 - 4 + 32 = 0
\]
Simplifying gives:
\[
(x - 8)^2 + (y - 2)^2 - 36 = 0
\]
\[
(x - 8)^2 + (y - 2)^2 = 36
\]
This represents a circle centered at \((8, 2)\) with a radius of \(6\).
4. To express the equation in the slope-intercept form (\(y = mx + b\)), we can derive the function by solving for \(y\):
\[
(y - 2)^2 = 36 - (x - 8)^2
\]
Taking the square root gives:
\[
y - 2 = \pm \sqrt{36 - (x - 8)^2}
\]
So we have:
\[
y = 2 \pm \sqrt{36 - (x - 8)^2}
\]
This isn't strictly slope-intercept form since the square root would not lead to a linear equation. Instead, you'd typically represent both branches of the circle or focus on lines of tangency if looking for linear approximations.
The key takeaway is that this equation represents a circle, so it cannot be expressed simply in the slope-intercept form like a line. For straight-line equations, the slope-intercept form would be applicable, but for curves like circles, you'll need to analyze them differently depending on the context. If you need specifically linear segments or tangents, further steps would be required!
1. Rearranging the equation:
\[
y^2 - 4y + x^2 - 16x + 32 = 0
\]
2. We will complete the square for both \(x\) and \(y\).
**Completing the square for \(x\)**:
- The expression \(x^2 - 16x\) can be completed as follows:
\[
x^2 - 16x = (x - 8)^2 - 64
\]
**Completing the square for \(y\)**:
- The expression \(y^2 - 4y\) can be completed as follows:
\[
y^2 - 4y = (y - 2)^2 - 4
\]
3. Now substituting back into the equation:
\[
(x - 8)^2 - 64 + (y - 2)^2 - 4 + 32 = 0
\]
Simplifying gives:
\[
(x - 8)^2 + (y - 2)^2 - 36 = 0
\]
\[
(x - 8)^2 + (y - 2)^2 = 36
\]
This represents a circle centered at \((8, 2)\) with a radius of \(6\).
4. To express the equation in the slope-intercept form (\(y = mx + b\)), we can derive the function by solving for \(y\):
\[
(y - 2)^2 = 36 - (x - 8)^2
\]
Taking the square root gives:
\[
y - 2 = \pm \sqrt{36 - (x - 8)^2}
\]
So we have:
\[
y = 2 \pm \sqrt{36 - (x - 8)^2}
\]
This isn't strictly slope-intercept form since the square root would not lead to a linear equation. Instead, you'd typically represent both branches of the circle or focus on lines of tangency if looking for linear approximations.
The key takeaway is that this equation represents a circle, so it cannot be expressed simply in the slope-intercept form like a line. For straight-line equations, the slope-intercept form would be applicable, but for curves like circles, you'll need to analyze them differently depending on the context. If you need specifically linear segments or tangents, further steps would be required!