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If $\mathrm{L}_1, \mathrm{~L}_2$ and $\mathrm{L}_3$ are the chords of contact of the three points $(2,0),(1,-2)$ and $(4,4)$ respectively with respect to the circle $x^2+y^2=3$, then $\mathrm{L}_1, \mathrm{~L}_2, \mathrm{~L}_3$ are
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Verified Answer
The correct answer is:
concurrent lines
Given equation of circle $x^2+y^2-3=0$
$\therefore$ chord of contacts of points $(2,0),(1,-2)$ and $(4,4)$ are
$$
\begin{aligned}
& \mathrm{L}_1: 2 \mathrm{x}+\mathrm{y} \cdot 0-3=0 \Rightarrow \mathrm{L}_1: 2 \mathrm{x}-3=0 \\
& \mathrm{~L}_2: 1 \cdot \mathrm{x}-2 \mathrm{y}-3=0 \Rightarrow \mathrm{L}_2: \mathrm{x}-2 \mathrm{y}-3=0 \\
& \mathrm{~L}_3: 4 \mathrm{x}+4 \mathrm{y}-3=0
\end{aligned}
$$
From(i) and(ii), we get $x=\frac{3}{2}$ and $y=\frac{-3}{4}$
Putting in (iii)
$4\left(\frac{3}{2}\right)+4\left(-\frac{3}{4}\right)-3=0$
$\Rightarrow 0=0$ (which satisfy equation (iii))
So, $\mathrm{L}_1, \mathrm{~L}_2$ and $\mathrm{L}_3$ are concurrent lines.
$\therefore$ chord of contacts of points $(2,0),(1,-2)$ and $(4,4)$ are
$$
\begin{aligned}
& \mathrm{L}_1: 2 \mathrm{x}+\mathrm{y} \cdot 0-3=0 \Rightarrow \mathrm{L}_1: 2 \mathrm{x}-3=0 \\
& \mathrm{~L}_2: 1 \cdot \mathrm{x}-2 \mathrm{y}-3=0 \Rightarrow \mathrm{L}_2: \mathrm{x}-2 \mathrm{y}-3=0 \\
& \mathrm{~L}_3: 4 \mathrm{x}+4 \mathrm{y}-3=0
\end{aligned}
$$
From(i) and(ii), we get $x=\frac{3}{2}$ and $y=\frac{-3}{4}$
Putting in (iii)
$4\left(\frac{3}{2}\right)+4\left(-\frac{3}{4}\right)-3=0$
$\Rightarrow 0=0$ (which satisfy equation (iii))
So, $\mathrm{L}_1, \mathrm{~L}_2$ and $\mathrm{L}_3$ are concurrent lines.
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