The equation of the circle with diameter endpoints of (13, -1) and (-15,9)
is: [tex](x+1)^2 + (y-4)^2 = 221[/tex]
If a circle O has radius of r units length and that it has got its center positioned at (h, k) point of the coordinate plane, then, its equation is given as:
[tex](x-h)^2 + (y-h)^2 = r^2[/tex]
Suppose we've two endpoints of a line segment as:
A(p,q), and B(m,n)
Then let the midpoint be M(x,y) on that line segment.
Then, its coordinates are:
[tex]x = \dfrac{p+m}{2}[/tex] (half of sum of x-coordinates of A and B) and
[tex]y = \dfrac{q+n}{2}[/tex] (half of sum of y-coordinates of A and B)
The shortest distance(length of the straight line segment's length connecting both given points) between points ( p,q) and (x,y) is:
[tex]D = \sqrt{(x-p)^2 + (y-q)^2} \: \rm units.[/tex]
For the considered case, a diameter of the considered circle has endpoints on (13,-1) and (-15,9).
Let we name those points as:
A(13,-1), and B(-15,9)
Let the center be O(x,y)
Then, the OA and OB should have same length and thus, O is the midpoint of AB.
Its coordinates will be:
[tex]x = \dfrac{13 + (-15)}{2} = -1\\\\y =\dfrac{-1 + 9}{2} = 4[/tex]
Thus, O(-1, 4) is the center of the circle.
The length of the radius of the circle is length of OA or OB (both same).
We get this distance as:
[tex]r = |OA| = \sqrt{(-1 - 13)^2 + (4-(-1))^2} =\sqrt{221}[/tex]
Thus, we get the equation of the circle as:
[tex](x-h)^2 + (y-h)^2 = r^2[/tex]
[tex](x-(-1))^2 + (y-4)^2 = (\sqrt{221})^2\\(x+1)^2 + (y-4)^2 = 221[/tex]
Thus, the equation of the circle with diameter endpoints of (13, -1) and (-15,9)
is: [tex](x+1)^2 + (y-4)^2 = 221[/tex]
Learn more about distance between two points here:
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