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(A) Let $x_1$ and $x_2$ be the roots of the equation $x^2+2x-a^2=0$. Thus,$(x-x_1)(x-x_2) = x^2+2x-a^2 = 0$.Similarly,let $y_1$ and $y_2$ be the roots

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$x^2+y^2+2x+4y-a^2-b^2=0$

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(A) Let $x_1$ and $x_2$ be the roots of the equation $x^2+2x-a^2=0$. Thus,$(x-x_1)(x-x_2) = x^2+2x-a^2 = 0$.Similarly,let $y_1$ and $y_2$ be the roots of the equation $y^2+4y-b^2=0$. Thus,$(y-y_1)(y-y_2) = y^2+4y-b^2 = 0$.Let the coordinates of points $A$ and $B$ be $(x_1, y_1)$ and $(x_2, y_2)$ respectively.The equation of a circle with diameter $AB$ is given by $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.Substituting the given quadratic expressions:$(x^2+2x-a^2) + (y^2+4y-b^2) = 0$.Rearranging the terms,we get:$x^2+y^2+2x+4y-a^2-b^2=0$.

Suppose that the $x$-coordinates of the points $A$ and $B$ satisfy $x^2+2x-a^2=0$ and their $y$-coordinates satisfy $y^2+4y-b^2=0$. Then,the equation of the circle with $AB$ as its diameter is

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