Question 1043513
{{{drawing(700,400,-3.5,3.5,-0.5,3.5,
grid(0),green(arrow(0,0,2,4sqrt(2))),
circle(0,0,3),circle(0.5,sqrt(2),1.5),
circle(0,0,0.05),circle(0.5,sqrt(2),0.05),
green(triangle(0,0,0.5,0,0.5,sqrt(2))),
green(rectangle(0.5,0,0.4,0.1)),
green(arrow(0.5,-1,0.5,5)),
locate(0.08,0.17,O),locate(0.57,1.47,C),
locate(0.55,0.17,D),locate(1.05,2.95,B),
locate(0.95,0.17,A),locate(-2.13,2.13,x^2+y^2=9)
)}}}
Since the new circle you are looking for passes through {{{O(0,0)}}} and {{{A(1,0)}}} ,
those two points are equidistant from the center {{{C}}} of the new circle,
and that means that center is on the bisectrix of segment {{{OA}}} .
That bisectrix is the line {{{CD}}} , with equation {{{x=0.5}}} .
If the new circle touches (meaning it is tangent to the circle {{{x^2+y^2=9}}} with radius {{{3}}} ,
then the radii of the two circles at the point of tangency {{{B}}} are on the same line {{{OC}}} .
So, {{{OB=3}}} is a radius of circle {{{x^2+y^2=3}}} ,
and a diameter of the new circle,
and {{{OB=2*OC=3}}} ,
meaning that {{{OC=CB=3/2=1.5}}} .
Applying the Pythagorean theorem to right triangle {{{OCD}}} , we have
{{{CD^2+0.5^2=1.5^2}}}
{{{CD^2=1.5^2-0.5^2
{{{CD^2=2}}
{{{CD=sqrt(2)}}} .
OF course, there are two possibilities:
{{{C(0.5,sqrt(2))}}} as in the drawing above, with the new circle mostly above the x-axis, or
{{{C(0.5,-sqrt(2))}}} if the new circle "hangs mostly below the x-axis."