Find the equation of a circle with diameter endpoints of (5,1)and(3,-1)

Answer 1

The equation of a circle is (x-h)² + (y-k)² = r², where (h,k) is the center of the circle and "r" is the radius. The center point of the circle will be the midpoint of the diameter. The endpoints of the diameter are (5,1) and (3,-1) and the midpoint, (h,k), is found by ((5+3)/2,(1-1)/2) = (8/2,0/2) = (4,0). Therefore, h = 4 and k = 0. To find the radius, r, we can find the distance from (4,0) to (5,1) by √((x₂ - x₁)² + (y₂ - y₁)²):

r = √((5-4)² + (1 - 0)²)

r = √(1² + 1²)

r = √2

Put these values, h = 4, k = 0 and r = √2, into the formula for a circle we get:

(x-h)² + (y-k)² = r²

(x-4)² + (y-0)² = (√2)²

(x-4)² + y² = 2

Hope this helps!

Answer 2

Diameter is twice the radius.

the distance between the two points is the square root of the sum of squares of the difference between the coordinates: d = sqr (5-3)^2 + (-1-1)^2 = sqr(4+4) = 2sqr2 radius = d/2 = sqr2

The general formula for a circle is (x-h)^2 + (y-k)^2 = r^2 where (h,k) is the circle's center

plug in r=sqr2 and r^2 =2

The center is the midpoint between the two end points, just average the x and y coordinates

Center is (4,0) = (h,k)

so the circle equation is (x-4)^2 + y^2 = 2

check and plug in each point to see if it satisfies the equation

(5,1) gives 1 + 1 = 2

(3,-1) gives 1 + 1 = 2