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$(a, 0)$ and $(b, 0)$ are centres of two circles belonging to a coaxial system of which $y$-axis is the radical axis. If radius of one of the circles is ' $r$ ', then the radius of the other circle is
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Verified Answer
The correct answer is:
$\left(r^2+b^2-a^2\right)^{1 / 2}$
Let the equation of circle whose centre $(a, 0)$ and radius $(r)$ is
$$
\begin{gathered}
(x-a)^2+(y-0)^2=r^2 \\
\Rightarrow \quad S_1 \equiv x^2+a^2-2 a x+y^2-r^2=0
\end{gathered}
$$
and the equation of circle whose centre $(b, 0)$ and radius $R$ is
$$
\begin{gathered}
(x-b)^2+(y-0)^2=R^2 \\
\Rightarrow \quad S_2 \equiv x^2+b^2-2 b x+y^2-R^2=0
\end{gathered}
$$
$\therefore$ Equation of radical axis is
$$
\begin{gathered}
S_1-S_2=0 \\
\Rightarrow \quad a^2-b^2+2 b x-2 a x+R^2-r^2=0 \\
\Rightarrow \quad R^2=r^2-a^2+b^2-2 b x+2 a x
\end{gathered}
$$
Since, radical axis is $y$-axis.
Therefore, putting $x=0$ in Eq. (i), we get
$$
\begin{aligned}
& R^2=r^2-a^2+b^2-0+0 \\
\Rightarrow \quad & R=\left(r^2+b^2-a^2\right)^{1 / 2}
\end{aligned}
$$
$$
\begin{gathered}
(x-a)^2+(y-0)^2=r^2 \\
\Rightarrow \quad S_1 \equiv x^2+a^2-2 a x+y^2-r^2=0
\end{gathered}
$$
and the equation of circle whose centre $(b, 0)$ and radius $R$ is
$$
\begin{gathered}
(x-b)^2+(y-0)^2=R^2 \\
\Rightarrow \quad S_2 \equiv x^2+b^2-2 b x+y^2-R^2=0
\end{gathered}
$$
$\therefore$ Equation of radical axis is
$$
\begin{gathered}
S_1-S_2=0 \\
\Rightarrow \quad a^2-b^2+2 b x-2 a x+R^2-r^2=0 \\
\Rightarrow \quad R^2=r^2-a^2+b^2-2 b x+2 a x
\end{gathered}
$$
Since, radical axis is $y$-axis.
Therefore, putting $x=0$ in Eq. (i), we get
$$
\begin{aligned}
& R^2=r^2-a^2+b^2-0+0 \\
\Rightarrow \quad & R=\left(r^2+b^2-a^2\right)^{1 / 2}
\end{aligned}
$$
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