Question 1131587
x^2+y^2=9 and to x=5. thank you
P.s Is my answer y^2=-4(x-1) and y^2=-16(x-4) 
<pre>
{{{drawing(400,400,-4,6,-4,6, 

locate(2.3,3.15,"(x,y)"),green(line(55/16,3,5,3)),
locate(5,3.15,"(5,y)"),blue(line(165/73,144/73)),red(line(165/73,144/73,55/16,3)),
graph(400,400,-4,6,-4,6), red(circle(0,0,3)), circle(55/16,3,25/16), line(5,-11,5,11)


)}}} 

The larger red circle is x²+y²=9.  The smaller black circle is a typical 
circle tangent to the large circle and to the vertical line x=5. Its center
is a point (x,y) on the desired equation. 

[blue radius] = 3

[green radius] = 5-x = [red radius]

[blue radius] + [red radius] = [distance from (0,0) to (x,y)] = &#8730;<span style="text-decoration: overline">x²+y²</span>)

So the un-simplified equation is:

3 + (5-x) = &#8730;<span style="text-decoration: overline">x²+y²</span>  

8 - x = &#8730;<span style="text-decoration: overline">x²+y²</span>

Squaring both sides:

(8 - x)² = (&#8730;<span style="text-decoration: overline">x²+y²</span>)²

64 - 16x + x² = x² + y²

64 - 16x = y²

y² = -16x + 64

y² = -16(x - 4),  a parabola

You got that for one of your answers.

[Where did you get y² = -4(x-1)?]

Now we put that parabola on the other one graph, and draw a couple more
circles, using a different scale:

{{{drawing(400,400,-10,10,-10,10,
circle(55/16,3,.05),line(5,-10,5,10),
graph(400,400,-10,10,-10,10), red(circle(0,0,3)), circle(55/16,3,25/16),
graph(400,400,-10,10,-10,10,-4sqrt(4-x)),graph(400,400,-10,10,-10,10,4sqrt(4-x)),
circle(3,4,2),circle(4,0,1),
circle(0,8,5), circle(-.75,-8.7178,5.75), circle(-.5,8.4853,5.5),
circle(-3,10.583,8),circle(1,0,4)


)}}}

Edwin</pre>