Find the standard form for the equation of a circle
[tex](x-h)^2+(y-k)^2=r^2[/tex]
with a diameter that has endpoints (-6,9) and (5,8).
[tex]
\begin{array}{l}
h= \\
k= \\
r=
\end{array}
[/tex]
Answer (1)
Find the center ( h , k ) of the circle by calculating the midpoint of the diameter using the midpoint formula: h = 2 − 6 + 5 = − 0.5 and k = 2 9 + 8 = 8.5 .
Calculate the diameter d using the distance formula: d = ( 5 − ( − 6 ) ) 2 + ( 8 − 9 ) 2 = 122 . Then, find the radius r by dividing the diameter by 2: r = 2 122 .
Substitute h , k , and r into the standard form equation of a circle: ( x − h ) 2 + ( y − k ) 2 = r 2 , which gives ( x + 0.5 ) 2 + ( y − 8.5 ) 2 = ( 2 122 ) 2 .
Simplify the equation to get the final standard form: ( x + 0.5 ) 2 + ( y − 8.5 ) 2 = 30.5 .
Explanation
Problem Analysis We are given the endpoints of a diameter of a circle as ( − 6 , 9 ) and ( 5 , 8 ) . Our goal is to find the standard form equation of the circle, which is ( x − h ) 2 + ( y − k ) 2 = r 2 , where ( h , k ) is the center and r is the radius.
Finding the Center First, we need to find the center of the circle ( h , k ) . The center is the midpoint of the diameter. We use the midpoint formula: ( 2 x 1 + x 2 , 2 y 1 + y 2 ) .
So, h = 2 − 6 + 5 = 2 − 1 = − 0.5 and k = 2 9 + 8 = 2 17 = 8.5 .
Finding the Radius Next, we need to find the radius r . The radius is half the length of the diameter. We first find the length of the diameter using the distance formula: d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 .
So, d = ( 5 − ( − 6 ) ) 2 + ( 8 − 9 ) 2 = ( 5 + 6 ) 2 + ( 8 − 9 ) 2 = 1 1 2 + ( − 1 ) 2 = 121 + 1 = 122 .
Then, the radius is r = 2 d = 2 122 .
Standard Form Equation Now, we substitute the values of h , k , and r into the standard form equation of a circle: ( x − h ) 2 + ( y − k ) 2 = r 2 .
( x − ( − 0.5 ) ) 2 + ( y − 8.5 ) 2 = ( 2 122 ) 2
( x + 0.5 ) 2 + ( y − 8.5 ) 2 = 4 122
( x + 0.5 ) 2 + ( y − 8.5 ) 2 = 30.5
Final Answer Therefore, the standard form equation of the circle is ( x + 0.5 ) 2 + ( y − 8.5 ) 2 = 30.5 .
h = − 0.5 k = 8.5 r = 2 122 ≈ 5.52
Examples
Circles are fundamental in many real-world applications, from designing gears and wheels in mechanical engineering to modeling the orbits of planets in astronomy. Understanding the equation of a circle allows engineers to calculate the dimensions and placement of circular components in machines, ensuring they function correctly. In architecture, circles are used in the design of domes, arches, and circular windows, requiring precise calculations to ensure structural integrity and aesthetic appeal. For example, knowing the endpoints of a circular window's diameter allows architects to determine its center and radius, which are crucial for its construction.
