Find the equation of the circle with diameter end points (3,2) and (-3,-4).?

Question

This is a Pre-calculus problem. please show work. Thank you

Answer 1

the diameter of the circle is
d^2 =(3-(-3))^2+(2-(-4))^2
=> d^2 = (3+3)^2+(2+4)^2
=> d^2 = 36+64 = 100
=> d = 100^(1/2) = 10

So the radius of the circle is
r = d/2 = 10/2 = 5

Centre of the circle is ((3+(-3))/2,(2+(-4))/2) = (0,-1)

hence the equation of the circle is
(x-0)^2 + (y-(-1))^2 = 5^2
=> x^2 + y^2 + 1 - 2.y.(-1) = 25
=> x^2 + y^2 +2y -24 = 0 --------------------Answer

Answer 2

The center will be halfway between the two points:
((3 - 3)/2, (-4 + 2)/2) = (0, -1)
The square of the radius will be (3^2 + 3^2) = 18
The equation of the circle, then is
(x - 0)^2 + (y - 3)^2 = 18
or
(x)^2 + (y - 3)^2 = 18

Answer 3

The equation of a circle is: (x - h)² + (y - ok)² = r² the place (h,ok) is the vertex and r is the radius. a million. Use the area formulation: D = ?((x2 - x1)² + (y2 - y1)²) to discover the radius, then purely plug the numbers interior the equation. 2. Use "polishing off the sq." recommendations with a view to alter the equation to time-honored sort. 3. Use the midpoint formulation: M = ((x1 + x2)/2, (y1 + y2)/2) to discover the midsection, then use the area formulation to discover the radius, the two by potential of dividing the diameter by potential of two or using the midpoint and an endpoint.