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If $(4,2)$ and $(k,-3)$ are conjugate points with respect to $x^2+y^2-5 x+8 y+6=0$, then $k$ equals
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Verified Answer
The correct answer is:
$\frac{28}{3}$
Equation of polar of point $(4,2)$ w.r.t. circle
$$
\begin{aligned}
& x^2+y^2-5 x+8 y+6=0 \text { is } T=0 \text {. } \\
& \therefore 4 x+2 y-\frac{5}{2}(x+4)+4(y+2)+6=0 \\
& \Rightarrow \quad 3 x+12 y+8=0 \\
&
\end{aligned}
$$
Since, points $(4,2)$ and $(k,-3)$ are conjugate points w.r.t. the given circle.
So, $(k,-3)$ satisfied line (i).
$$
\begin{array}{rlrl}
& \therefore & 3 k-36+8 & =0 \\
\Rightarrow & k & =\frac{28}{3}
\end{array}
$$
$$
\begin{aligned}
& x^2+y^2-5 x+8 y+6=0 \text { is } T=0 \text {. } \\
& \therefore 4 x+2 y-\frac{5}{2}(x+4)+4(y+2)+6=0 \\
& \Rightarrow \quad 3 x+12 y+8=0 \\
&
\end{aligned}
$$
Since, points $(4,2)$ and $(k,-3)$ are conjugate points w.r.t. the given circle.
So, $(k,-3)$ satisfied line (i).
$$
\begin{array}{rlrl}
& \therefore & 3 k-36+8 & =0 \\
\Rightarrow & k & =\frac{28}{3}
\end{array}
$$
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