Question 934410
The general equation of a circle is,
{{{(x-h)^2+(y-k)^2=R^2}}}
{{{(x-2)^2+(y-3)^2=R^2}}}
Use A to get R.
{{{(-2-2)^2+(0-3)^2=R^2}}}
{{{16+9=R^2}}}
{{{R^2=25}}}
{{{R=5}}}
.
.
.
{{{(x-2)^2+(y-3)^2=25}}}
{{{(5-2)^2+(k-3)^2=25}}}
{{{9+(k-3)^2=25}}}
{{{(k-3)^2=16}}}
{{{k-3= 0 +- 4}}}
{{{k=3 +- 4}}}
{{{k=-1}}} and {{{k=7}}}
.
.
.

{{{2(x-2)dx+2(y-3)dy=0}}}
{{{(y-3)dy=-(x-2)dx}}}
{{{dy/dx=-(x-2)/(y-3)}}}
At A,
{{{dy/dx=-(-2-2)/(0-3)}}}
{{{dy/dx=-(4/3)}}}
{{{drawing(300,300,-5,5,-5,5,grid(1),blue(line(-2,0,2,-5)),circle(2,3,5))}}}
The value of the derivative is equal to the slope of the tangent line.
Using the point-slope form of a line,
{{{y-0=-(4/3)(x-(-2))}}}
{{{y=-(4/3)(x+2)}}}
So when {{{x=0}}}
{{{y=-(4/3)(2)}}}
{{{y=-8/3}}}
So know you know the base and height of the triangle.
{{{A=(1/2)(2)(8/3)}}}
{{{A=8/3}}}