(a, 0) and (b, 0) are the centres of two circ | vedclass

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(B) Let the equation of the circle with centre $(a, 0)$ and radius $r$ be $(x-a)^2 + y^2 = r^2$,which simplifies to $S_1 \equiv x^2 + y^2 - 2ax + a^2

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$(r^2+b^2-a^2)^{1/2}$

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(B) Let the equation of the circle with centre $(a, 0)$ and radius $r$ be $(x-a)^2 + y^2 = r^2$,which simplifies to $S_1 \equiv x^2 + y^2 - 2ax + a^2 - r^2 = 0$. Let the equation of the circle with centre $(b, 0)$ and radius $R$ be $(x-b)^2 + y^2 = R^2$,which simplifies to $S_2 \equiv x^2 + y^2 - 2bx + b^2 - R^2 = 0$. The radical axis of the two circles is given by $S_1 - S_2 = 0$. $(x^2 + y^2 - 2ax + a^2 - r^2) - (x^2 + y^2 - 2bx + b^2 - R^2) = 0$. $-2ax + 2bx + a^2 - b^2 - r^2 + R^2 = 0$. $2x(b-a) + (a^2 - b^2 - r^2 + R^2) = 0$. Since the radical axis is the $y$-axis,its equation is $x = 0$. Comparing $2x(b-a) + (a^2 - b^2 - r^2 + R^2) = 0$ with $x = 0$,the constant term must be zero: $a^2 - b^2 - r^2 + R^2 = 0$. $R^2 = r^2 + b^2 - a^2$. $R = (r^2 + b^2 - a^2)^{1/2}$.

$(a, 0)$ and $(b, 0)$ are the centres of two circles belonging to a coaxial system of which the $y$-axis is the radical axis. If the radius of one of the circles is $r$,then the radius of the other circle is

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