If AC is the diagonal, and lies on the x-axis, we know that point C = (-5,0). In order to calculate the B point, we have to find the next intersection between the line and circle:

\(x^2 + y^2 = 25\)
\(x^2 + (3x+15)^2 = 25\)
\(10x^2 + 50x + 200 = 0\)
\(x^2 + 5x + 20 = 0\)

Then, we know that the roots are:
\(x=-4 or x =-5\)

We already know that -5 is one of the points, therefore, we have to evaluate -4.

Evaluating x=-4 in y=3x+15 yields y=3.

Having said that, the side BC is \(1^2 + 3^2 = BC^2\), \(BC = \sqrt{10}\)

Now, we can use the pythagorean theorem in order to calculate side AB. \(AB^2 + \sqrt{10}^2 = 10^2\), \(AB = \sqrt{90}\).

Finally, the area is
\(A = \sqrt{90}\sqrt{10}\)
\(A = \sqrt{900}\)
\(A = 30\)

Option C.