Question

Please help ASAP 60 pts!

Write the equation of a circle with endpoints of the diameter at (4, -3) and (-2, 5).
Show your work for credit.

Answer 1

• (x - 1)² + (y - 1)² = 25

Explanation:

Find the center, the midpoint of the diameter:

• x = (4 - 2)/2 = 1, this is h of the circle equation
• y = (-3 + 5)/2 = 1, his is k of the circle equation

Find the square of the radius by distance formula:

• r² = (1 - 4)² + (1 + 3)² = 25

Equation of the circle:

• (x - h)² + (y - k)² = r²,

Substitute the values to get:

• (x - 1)² + (y - 1)² = 25

Answer 2

solution given:

points at (4, -3) and (-2, 5

diameter [d]=

`d = \sqrt{(4 + 2) {}^(2) +( - 3 - 5) {}^(2) } \\ d = 10units \\ radius(r) = 5units`

centre[h,k]={(4-2)/2,(-3+5)/2}=(1,1)

now equation of a circle

(x-h)²+(y-k)²=r²

(x-1)²+(y-1)²=25 is a required equation