Answer:
Equation = x²+(y+3)²=5²
Step-by-step explanation:
The question is on equation of a circle in a center-radius form
The general equation is
(x-h)² + (y-k)² = r² where the center is at (h,k) and the radius is r
Given
point P=(3,1) and point Q= (-3,-7)
Find the diameter PQ
[tex]d=\sqrt{(X2-X1)^2 + (Y2-Y1)^2\\[/tex]
[tex]d= \sqrt{(-7-1)^2 + (-3-3)^2}[/tex]
[tex]d=\sqrt{-8^2 + -6^2}[/tex]
[tex]d= \sqrt{64+36} = 10[/tex]
d= 10 units hence radius is 10/2 =5 units
Find the center of the circle
center =[ (x₁+x₂)/2, (y₁+y₂)/2]
center=[ (3+-3)/2 , (1+-7)/2]
center= [ 0/2 ,-6/2]
center= [0,-3]
Equation = x²+(y+3)²=5²
