Question
A circle has a diameter with endpoints at (1, 0) and (−3, 10) . Rudy says the equation of the circle is (x+1)2+(y−5)2=29−−√ . Is Rudy correct?(1 point)ResponsesNo, his equation has an error. The equation should be (x−1)2+(y+5)2=29 because the center point of the circle is at (−1, 5) and because the term in the equation is r2, so he needs to square the radius.No, his equation has an error. The equation should be open paren x minus 1 close paren squared plus open paren y plus 5 close paren squared is equal to 29 because the center point of the circle is at (−1, 5) and because the term in the equation is r squared, so he needs to square the radius.No, his equation has an error. The equation should be (x+1)2+(y−5)2=29 because the term in the equation is r2, so he needs to square the radius.No, his equation has an error. The equation should be open paren x plus 1 close paren squared plus open paren y minus 5 close paren squared is equal to 29 because the term in the equation is r squared, so he needs to square the radius.Yes, Rudy's equation is correct.Yes, Rudy's equation is correct.No, his equation has an error. The equation should be (x−1)2+(y+5)2=29−−√ because the center point of the circle is at (−1, 5).
Answers
GPT-4o mini
To determine if Rudy's equation for the circle is correct, let's start by finding the center and radius of the circle based on the endpoints of the diameter.
1. The endpoints of the diameter are (1, 0) and (−3, 10).
2. The center of the circle \((h, k)\) is the midpoint of the diameter. The midpoint formula is:
\[
h = \frac{x_1 + x_2}{2}, \quad k = \frac{y_1 + y_2}{2}
\]
Plugging in the points:
\[
h = \frac{1 + (-3)}{2} = \frac{-2}{2} = -1
\]
\[
k = \frac{0 + 10}{2} = \frac{10}{2} = 5
\]
So the center is \((-1, 5)\).
3. Next, we need to calculate the radius \(r\). The radius is half the length of the diameter. First, we find the length of the diameter using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Plugging in our points:
\[
d = \sqrt{(−3 - 1)^2 + (10 - 0)^2} = \sqrt{(−4)^2 + 10^2} = \sqrt{16 + 100} = \sqrt{116} = 2\sqrt{29}
\]
Therefore, the radius \(r\) is:
\[
r = \frac{d}{2} = \frac{2\sqrt{29}}{2} = \sqrt{29}
\]
4. The standard equation of a circle is given by:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
Substituting the center and radius:
\[
(x + 1)^2 + (y - 5)^2 = (\sqrt{29})^2
\]
which simplifies to:
\[
(x + 1)^2 + (y - 5)^2 = 29
\]
Rudy's equation was:
\[
(x + 1)^2 + (y - 5)^2 = \sqrt{29}
\]
This equation is incorrect because it should be:
\[
(x + 1)^2 + (y - 5)^2 = 29
\]
So, the correct response is:
**No, his equation has an error. The equation should be (x + 1)² + (y - 5)² = 29 because the term in the equation is r² (not r), so he needs to square the radius.**
1. The endpoints of the diameter are (1, 0) and (−3, 10).
2. The center of the circle \((h, k)\) is the midpoint of the diameter. The midpoint formula is:
\[
h = \frac{x_1 + x_2}{2}, \quad k = \frac{y_1 + y_2}{2}
\]
Plugging in the points:
\[
h = \frac{1 + (-3)}{2} = \frac{-2}{2} = -1
\]
\[
k = \frac{0 + 10}{2} = \frac{10}{2} = 5
\]
So the center is \((-1, 5)\).
3. Next, we need to calculate the radius \(r\). The radius is half the length of the diameter. First, we find the length of the diameter using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Plugging in our points:
\[
d = \sqrt{(−3 - 1)^2 + (10 - 0)^2} = \sqrt{(−4)^2 + 10^2} = \sqrt{16 + 100} = \sqrt{116} = 2\sqrt{29}
\]
Therefore, the radius \(r\) is:
\[
r = \frac{d}{2} = \frac{2\sqrt{29}}{2} = \sqrt{29}
\]
4. The standard equation of a circle is given by:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
Substituting the center and radius:
\[
(x + 1)^2 + (y - 5)^2 = (\sqrt{29})^2
\]
which simplifies to:
\[
(x + 1)^2 + (y - 5)^2 = 29
\]
Rudy's equation was:
\[
(x + 1)^2 + (y - 5)^2 = \sqrt{29}
\]
This equation is incorrect because it should be:
\[
(x + 1)^2 + (y - 5)^2 = 29
\]
So, the correct response is:
**No, his equation has an error. The equation should be (x + 1)² + (y - 5)² = 29 because the term in the equation is r² (not r), so he needs to square the radius.**