(a, 0) and (b, 0) are centers of two circles | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the equation of the circle with center $(a, 0)$ 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 center $(b, 0)$ 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 $S_1 - S_2 = 0$. Substituting the equations,we get $(x^2 + y^2 - 2ax + a^2 - r^2) - (x^2 + y^2 - 2bx + b^2 - R^2) = 0$. This simplifies to $2(b-a)x + a^2 - b^2 - r^2 + R^2 = 0$. Since the radical axis is the $y$-axis,its equation must be $x = 0$. For the equation $2(b-a)x + (a^2 - b^2 - r^2 + R^2) = 0$ to represent $x = 0$,the constant term must be zero. Therefore,$a^2 - b^2 - r^2 + R^2 = 0$. Solving for $R$,we get $R^2 = r^2 + b^2 - a^2$,which implies $R = (r^2 + b^2 - a^2)^{1/2}$.

$(a, 0)$ and $(b, 0)$ are centers 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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