Question 331946
Given the conditions, we know that h=-3. That is, since the circle is tangent to the y-axis, lies in the third quadrant, and the radius is 3, the  x-coordinate of the center must be at x=-3.

To find the y-coordinate, k, of the center use y=2x.

y=2(-3)=-6

The center coordinates are (-3,-6). The product is obviously (-6)(-3)=18.

The equation of the circle is {{{(x+3)^2+(y+6)^2=9}}}

To graph this on most calculators, we would have to solve this for y in terms of x.

If you have a graphing calculator, graph:

{{{-(sqrt(-x^2-6x)+6)}}} for the lower half of the circle

and

{{{sqrt(-x^2-6x)-6}}} for the uppr half.

Graph y=2x as well and you can see it pass through the center of the circle.

Casio calcualators have a nice Conics menu that graphs conics in terms of there equations without solving for y. It's a nice feature.