What is the equation of a circle whose diameter endpoints are at 10 -4 and 2 2 on the Cartesian plane and what are the perpendicular equations to its endpoints showing work?

Answer 1

Endpoints of diameter: (10, -4) and (2, 2)

Midpoint which is the center of the circle: (6, -1)

Distance from (10, -4) or (2, 2) to (6, -1) = 5 which is the radius of the circle

Equation of the circle: (x-6)^2 +(y+1)^2 = 25

Slope of radius: -3/4

Slope of perpendicular equations which will be parallel: 4/3

1st perpendicular equation: y--4 = 4/3(x-10) => 3y = 4x-52

2nd perpendicular equation: y-2 = 4/3(x-2) => 3y = 4x-2

Answer 2

The centre is [(10+2)/2, (-4+2)/2] = (6, -1)The radius is sqrt[(10-6)^2 + (-4+1)^2] = sqrt(4^2 + 3^2) = 5

Therefore, the equation of the circle is (x - 6)^2 + (y + 1)^2 = 25

It is not clear what you mean by "perpendicular equations to its endpoints".