A circle is tangent to the y axis and has aradius of three units. The center of the circle is in the third quadrant and lies on the graph...

Let  (h,k) be the centre of the circle. Since y  axis is the
tangent,  The normal from the tangent || to x axis passes through the centre of the circle. So h
=| r| , the radius of the circle, radius is given to be 3.

Therefore
h = r = |3 |.   h should be in 3rd quadrant . So h = -3.

Therefore
the centre (h,k) = (-3,k) lies on y-2x. So (-3,k) should satisfy y-2x =
0.

So k -2(-3) = 0.  Or k+5 = 0 . So k
=-5.

Therefore the centre is at (-3, -5) which in 3rd
quadrant.

Therefore the  equation of the is (x-h)^2 +(y-k) = r^2 .
Or

{x-(-3)}^2 +{y- (-5)}^2 =
3^2.

(x+3)^2+(x+5)^2 =
3^2.

x^2+6x+9+y^2+10y+25 =
9.

Rearranging the standard form, we get the equation of the
circle,

x^2+y^2+6x+10y +25 = 0 which has the centre at (-3, -5) and
radius  3 units, toucching y  axis.