You can do the problem without a special formula. A tangent line to a circle is perpendicular to the circle's radius at the tangent point. So if we take the first circle, with center (1, -4), and draw its radius to the tangent line, the slope of that radius will be -1 (since it is perpendicular to y = x + 3). If the slope is -1, the equation of that radius-line is y = -x + b, and since the center (1, -4) is on that line, we can substitute into the line's equation to find b: 1 = 4 + b, so b = -3. So the equation of the radius-line is y = -x - 3. If we solve for where this radius meets the tangent line, by solving the two equations for the lines together, we find they meet at (-3, 0). From (1, -4) to (-3, 0) we go across 4 units and up 4 units, so using Pythagoras, if r is the radius, r^2 = 4^2 + 4^2 = 32, and the area of the circle is 32π.
You could do the same for the second circle too, but if you notice the center of that circle, (5, -8), is also on the radius-line we just found, you can see immediately that it will also meet the tangent line at the same point, (-3, 0). Since we go across 8 units and up 8 units from (5, -8) to (-3, 0), then by Pythagoras, if R is the radius, R^2 = 8^2 + 8^2 = 128, and the area of the larger circle is 128π.
So the ratio of the two areas is 32 to 128, or 1 to 4.
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