Write equation in standard form, of the circle w/ center C(6,-2) and endpoints are at (6,0) and (6,-4)?

Use the endpoints of the circle to find the distance across the circle, the diameter.


sqrt [(6-6)^2 + (0- -4)^2)]


sqrt(0+16)


4, so the diameter is 4 which means the radius is 2.


Now using the center points


(x-6)^2 + (y+2)^2 = radius squared


(x-6)^2 + (y+2)^2 = 4

Write equation in standard form, of the circle w/ center C(6,-2) and endpoints are at (6,0) and (6,-4)?
clearly (6,0) and (6,-4) are the end points of a diameter





an alternative method to the other answers is to use the form





(x-6)(x-6) + (y-0)(y - (-4)) = 0
Reply:Graphing the center and endpoints, it is easy to deduct that the circle has a radius of 2.





Therefore, by applying the formula: (x - h)^2 + (y - k)^2 = r^2 you can write the circle's equation, h %26amp; k being the center's coordinates and r the radius.





(x - 6)^2 + (y + 2)^2 = 4





Foil the expression and simplify...





x^2 - 12x + 36 + y^2 + 4y + 4 = 4





x^2 + y^2 - 12x + 4y + 36 = 0
Reply:diameter =sqrt[(6-6)^2+(0+4)^2]=4


radius = 2


(x-6)^2+(y-(-2))^2=2^2


(x-6)^2+(y+2)^2=4


The equation of a circle with center (a,b) and radius r is


(x-a)^2+(y-b)^2=r^2