Question 1157854
<pre>
That means:

Find all points on the circle x^2 + y^2 = 81 where the slope OF A LINE DRAWN
TANGENT TO THE CIRCLE AT THAT POINT is 9/40.

{{{drawing(375,375,-12,12,-12,12, graph(375,375,-12,12,-12,12),
circle(0,0,9), line(-13,6.3,15,12.6), line(13,-6.3,-15,-12.6),
circle(-81/41,360/41,.2), circle(81/41,-360/41,.2),locate(-4.5,10,"(?,?)"),locate(2,-9,"(?,?)")  



 )}}}

{{{x^2 + y^2 = 81}}}

We take the derivative implicitly

{{{2x + 2y*expr(dy/dx) = 0}}}

Divide every term by 2

{{{x + y*expr(dy/dx) = 0}}}

Substitute 9/40 for (dy)/(dx)

{{{x + y*expr(9/40) = 0}}}

Solve for y:

{{{y*expr(9/40) = -x}}}

Multiply both sides by 40/9

{{{y = -40x/9}}}

Substitute for y in the equation of the circle:

{{{x^2 + (-40x/9)^2 = 81}}}

{{{x^2 + 1600x^2/81 = 81}}}

Multiply through by 81

{{{81x^2 + 1600x^2 = 6561}}}

{{{1681x^2=6561}}}

{{{x^2=6561/1681}}}

{{{x="" +- 81/41)}}}

Substituting for x² in the equation of the circle:

{{{x^2 + y^2 = 81}}}

{{{6561/1681 + y^2 = 81}}}

Multiply through by 1681

{{{6561+1681y^2=136161}}}

{{{1681y^2=129600}}}

{{{y^2=129600/1681}}}

{{{y="" +- 360/41}}}

{{{1681y^2=6480}}}

{{{y^2=6480/1681}}}

{{{y="" +- 360/41}}}

So the two points where the slope of the tangent line is 9/40 are:

{{{(matrix(1,3,-81/41,",",360/41)))}}}

and

{{{(matrix(1,3,81/41,",",-360/41)))}}}

Edwin</pre>