Acircle of radius 5 has its center in the first quadrant and is tangent to the x-axis at (1,0).
find the equation of the circle
find the y intercepts of the circle

Ok, this is kinda crazy, but I THINK I got it. If the circle is tangent to the point (1,0), that means that if the radius is a length of 5, from the y coordinate of 0, you are going up 5 units to meet the center of the circle. That puts the vertex at that x coordinate which is 1, and the y coordinate which is 5. So using that info, the equation for the circle is (x-1)^2 + (y-5)^2 = 25. Now to find the y intercepts, we set x equal to 0: (0-1)^2 + (y-5)^2 = 25. -1 squared is 1. Subtract that from both sides to get (y-5)^2 = 24. Now, undo that square by taking the square root of both sides (using your calculator) to get
y-5= +/-4.89898.  Now add 5 to both sides to get y values of 9.89898 and .10102

128/3reason being 5 and 1/3 = 16/316/3 • 2/3 = 32/31 and 1/3 = 4/3 32/3 • 4/3 this will equal 128/3