I asked this yesterday and someone answered. but i didn't understand the explanation..
Two circles of radius 4 are tangent to the graph of y^2=4x at the point (1,2). Find equations of these two circles.
I got (x-1)^2+(y-2)^2=16 and
(x-4.36)^2+(y+1.36)^2=16
The center of the circles is not at (1,2), so your first equation is incorrect.
the line joining (1,2) to the centers of the circles will be perpendicular to the tangent line. So, since the tangent line has slope 1, the normal line will have slope -1.
Now we have a point and a slope. Since the radius from the circle center (h,k) has length 4, then the (h,k) is at (1-2√2,2+2√2) or (1+2√2,2-2√2).
Now we have (h,k) and r for each circle.
(x-(1-2√2))^2 + (y-(2+2√2))^2 = 16
(x-(1+2√2))^2 + (y-(2-2√2))^2 = 16
(x+1.8)^2 + (y-4.8)^2 = 16
(x-3.8)^2 + (y+0.8)^2 = 16
To see the graphs, visit wolframalpha.com and enter
plot y=3-x,y=2√x,(x-(1-2√2))^2 + (y-(2+2√2))^2 = 16, (x-(1+2√2))^2 + (y-(2-2√2))^2 = 16