Higgins Potter
07/02/2023 · Elementary School
Find the standard form of the equation of the circle with endpoints of a diameter at the points \( (1,6) \) and \( (-9,4) \). Type the standard form of the equation of this circle. (Type an equation.)
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Step-by-step Solution
To find the standard form of the equation of a circle given the endpoints of a diameter, we first need to determine the center and the radius of the circle.
1. **Find the center of the circle**: The center of the circle is the midpoint of the diameter. The midpoint \( M \) of the points \( (x_1, y_1) = (1, 6) \) and \( (x_2, y_2) = (-9, 4) \) can be calculated using the midpoint formula:
\[
M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\]
Substituting the coordinates:
\[
M = \left( \frac{1 + (-9)}{2}, \frac{6 + 4}{2} \right) = \left( \frac{-8}{2}, \frac{10}{2} \right) = (-4, 5)
\]
So, the center of the circle is \( (-4, 5) \).
2. **Find the radius of the circle**: The radius is half the distance between the endpoints of the diameter. We can find the distance \( d \) between the points \( (1, 6) \) and \( (-9, 4) \) using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Substituting the coordinates:
\[
d = \sqrt{((-9) - 1)^2 + (4 - 6)^2} = \sqrt{(-10)^2 + (-2)^2} = \sqrt{100 + 4} = \sqrt{104} = 2\sqrt{26}
\]
The radius \( r \) is half of this distance:
\[
r = \frac{d}{2} = \frac{2\sqrt{26}}{2} = \sqrt{26}
\]
3. **Write the standard form of the equation of the circle**: The standard form of the equation of a circle with center \( (h, k) \) and radius \( r \) is given by:
\[
(x - h)^2 + (y - k)^2 = r^2
\]
Substituting \( h = -4 \), \( k = 5 \), and \( r = \sqrt{26} \):
\[
(x - (-4))^2 + (y - 5)^2 = (\sqrt{26})^2
\]
This simplifies to:
\[
(x + 4)^2 + (y - 5)^2 = 26
\]
Thus, the standard form of the equation of the circle is:
\[
\boxed{(x + 4)^2 + (y - 5)^2 = 26}
\]
Quick Answer
\((x + 4)^2 + (y - 5)^2 = 26\)
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