The equation(s) of the circle(s) which pass through the ends of the common chords of two circles $2x^{2}+2y^{2}+8x+4y-7=0$ and $x^{2}+y^{2}-8x-4y-5=0$ and touch the line $x=7$ is (are) :

Find the common chord by subtracting the circle equations, giving 10x + 8y + 1 = 0, which is the radical axis x + (4/5)y + 1/10 = 0. Substitute x = 7 into the radical axis to find the intersection point is (7, -9/4), which acts as the point of contact for the required circle. The required circle must be a member of the coaxal family, written as S1 + k(Radical Axis) = 0: 2x^2 + 2y^2 + 8x + 4y - 7 + k(10x + 8y + 1) = 0. The center of this family circle is (-(8+10k)/4, -(4+8k)/4). Since the circle touches the line x = 7, the distance from the center to the line equals its radius, which results in k = -31/5. Substituting k = -31/5 into the family equation and simplifying yields the circle x^2 + y^2 - 6x + 2y - 19/4 = 0.

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