SOLUTION: Hi I've just started working on my conics unit. I really thought I understood how to get an equation from a circle and vise versa BUT then I started to work in my module and got th

Question 280420: Hi I've just started working on my conics unit. I really thought I understood how to get an equation from a circle and vise versa BUT then I started to work in my module and got this question. Determine the standard form of the equation of a circle with a center at(4,-1) and passing through(0,2). By looking at the center points I can write part of the equation. (x-4)^2 +(y+1)^2 but I have no radius or diagram of circle to get radius from so would it therefore be 1 or is this where the(0,2) comes in??? I'm not sure what or how I'm suppose to use the(0,2). Could you please explain and thanks so very much.
Found 2 solutions by richwmiller, jim_thompson5910:
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
the radius is the distance from the center to the circle edge
you have the center 4,-1 and a point on the circle 0,2
find the distance .
That is the radius

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Hint: Recall that the radius is the distance from the center of a circle to any point on the circle. So this means that the radius will be equal to the distance from the center (4, -1) to the point on the circle (0,2). So to find the radius, simply find the distance from (4, -1) to (0,2). Let me know if you need help with that.
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