The equation of the circle is (x+8)^2 +(y−3)^2 =64.
To determine the equation of the circle with a center at C(-8, 3) and tangent to the y-axis, we recall that the standard form of the equation of a circle is (x−h)^2 +(y−k)^2 =r^2, where (h, k) represents the center and r is the radius.
Since the circle is tangent to the y-axis, its radius is the horizontal distance from the center to the y-axis, which is 8 units (the x-coordinate of the center). Therefore, the radius (r) is 8.
Substitute the values into the standard form, we get (x+8)^2 +(y−3)^2 =8^2 , which simplifies to (x+8)^2 +(y−3)^2 =64. Hence, the equation of the circle satisfying the given conditions is (x+8)^2 +(y−3)^2 =64.