Question 1039233
Find the equation of the circle which touches the X axis at the point (3,0) and cuts off an intercept 8 units from the positive part of the y axis

The standard equation of any circle is

{{{(x-h)^2+(y-k)^2=r^2}}}

Since it is tangent to the x-axis at (3,0),
the center is directly above (3,0), so its
center also has x-coordinate 3. So h=3

{{{(x-3)^2+(y-k)^2=r^2}}}

It passes through (x,y) = (3,0), so

{{{(3-3)^2+(0-k)^2=r^2}}}

{{{0^2+k^2=r^2}}}

{{{k^2=r^2}}}

k and r are both positive numbers.

{{{k=r}}} and

{{{(x-3)^2+(y-r)^2=r^2}}}

It has a y-intercept at (0,8)

{{{(0-3)^2+(8-r)^2=r^2}}}

{{{(-3)^2+64-16r+r^2=r^2}}}

{{{9+64-16r+r^2=r^2}}}

{{{73-16r=0}}}

{{{-16r=-73}}}

{{{r=(-73)/(-16)}}}

{{{r=73/16}}}

And since {{{x=r=73/16}}},

{{{(x-3)^2+(y-73/16)^2=(73/16)^2}}}

{{{drawing(4800/13,400,-3,9, -3,10,circle(3, 73/16,.1),
locate(3,73/16,(matrix(1,3,3,",",73/16))),

graph(4800/13,400,-3,9, -3,10), circle(3,73/16,73/16) )}}}