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The length of the tangent drawn from any point on the circle \(x^2+y^2+2 g x+2 f y+c_1=0\) to the circle \(x^2+y^2+2 g x+2 f y+c_2=0\) is
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Verified Answer
The correct answer is:
\(\sqrt{C_2-c_1}\)
Circles are
\(\begin{aligned}
& C_1 \equiv x^2+y^2+2 g x+2 f y+c_1=0 \\
\text {and } & C_2 \equiv x^2+y^2+2 g x+2 f y+c_2=0
\end{aligned}\)
Clearly circles are concentric,
Clearly, length of tangent is
\(\begin{aligned}
A T & =\sqrt{A O^2-O T^2}=\sqrt{r_1^2-r_2^2} \\
& =\sqrt{\left(g^2+f^2-c_1\right)-\left(g^2+f^2-c_2\right)}=\sqrt{c_2-c_1}
\end{aligned}\)
\(\begin{aligned}
& C_1 \equiv x^2+y^2+2 g x+2 f y+c_1=0 \\
\text {and } & C_2 \equiv x^2+y^2+2 g x+2 f y+c_2=0
\end{aligned}\)
Clearly circles are concentric,
Clearly, length of tangent is
\(\begin{aligned}
A T & =\sqrt{A O^2-O T^2}=\sqrt{r_1^2-r_2^2} \\
& =\sqrt{\left(g^2+f^2-c_1\right)-\left(g^2+f^2-c_2\right)}=\sqrt{c_2-c_1}
\end{aligned}\)
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