Question
Find centre and radius of circle x²+y²-6x-8y-11=0
Answers
GPT-4o mini
To find the center and radius of the circle given by the equation \(x^2 + y^2 - 6x - 8y - 11 = 0\), we can rewrite the equation in the standard form of a circle, which is:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
where \((h, k)\) is the center of the circle and \(r\) is the radius.
1. Start by rearranging the equation:
\[
x^2 - 6x + y^2 - 8y - 11 = 0
\]
2. Next, we will complete the square for the \(x\) terms and the \(y\) terms.
**For the \(x\) terms:**
\[
x^2 - 6x
\]
To complete the square, take half of \(-6\), square it, and then add and subtract that value:
\[
\left(\frac{-6}{2}\right)^2 = 9
\]
Thus, we can rewrite the \(x\) terms as:
\[
(x^2 - 6x + 9) - 9
\]
which simplifies to:
\[
(x - 3)^2 - 9
\]
**For the \(y\) terms:**
\[
y^2 - 8y
\]
Similarly, take half of \(-8\), square it:
\[
\left(\frac{-8}{2}\right)^2 = 16
\]
So, the \(y\) terms become:
\[
(y^2 - 8y + 16) - 16
\]
which simplifies to:
\[
(y - 4)^2 - 16
\]
3. Now substitute these completed squares back into the equation:
\[
(x - 3)^2 - 9 + (y - 4)^2 - 16 - 11 = 0
\]
Combine the constant terms:
\[
(x - 3)^2 + (y - 4)^2 - 36 = 0
\]
4. Rearranging gives:
\[
(x - 3)^2 + (y - 4)^2 = 36
\]
5. From this equation, we can identify the center and the radius:
- **Center:** \((h, k) = (3, 4)\)
- **Radius:** \(r = \sqrt{36} = 6\)
Thus, the center of the circle is \((3, 4)\) and the radius is \(6\).
\[
(x - h)^2 + (y - k)^2 = r^2
\]
where \((h, k)\) is the center of the circle and \(r\) is the radius.
1. Start by rearranging the equation:
\[
x^2 - 6x + y^2 - 8y - 11 = 0
\]
2. Next, we will complete the square for the \(x\) terms and the \(y\) terms.
**For the \(x\) terms:**
\[
x^2 - 6x
\]
To complete the square, take half of \(-6\), square it, and then add and subtract that value:
\[
\left(\frac{-6}{2}\right)^2 = 9
\]
Thus, we can rewrite the \(x\) terms as:
\[
(x^2 - 6x + 9) - 9
\]
which simplifies to:
\[
(x - 3)^2 - 9
\]
**For the \(y\) terms:**
\[
y^2 - 8y
\]
Similarly, take half of \(-8\), square it:
\[
\left(\frac{-8}{2}\right)^2 = 16
\]
So, the \(y\) terms become:
\[
(y^2 - 8y + 16) - 16
\]
which simplifies to:
\[
(y - 4)^2 - 16
\]
3. Now substitute these completed squares back into the equation:
\[
(x - 3)^2 - 9 + (y - 4)^2 - 16 - 11 = 0
\]
Combine the constant terms:
\[
(x - 3)^2 + (y - 4)^2 - 36 = 0
\]
4. Rearranging gives:
\[
(x - 3)^2 + (y - 4)^2 = 36
\]
5. From this equation, we can identify the center and the radius:
- **Center:** \((h, k) = (3, 4)\)
- **Radius:** \(r = \sqrt{36} = 6\)
Thus, the center of the circle is \((3, 4)\) and the radius is \(6\).