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The locus of a point which is at a distance of 2 units from the line $2 x-3 y+4=0$ and at a distance of $\sqrt{13}$ units from a point $(5,0)$ is
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Verified Answer
The correct answer is:
$12 x y-5 y^2-56 x+24 y+84=0$
Let $P(h, k)$ be the point whose locus is to be find. given $\sqrt{(h-5)+(k-0)^2}=\sqrt{13}$
$$
\Rightarrow h^2=-k^2+10 h-12...(1)
$$
also given, $\frac{2 h-3 k+4}{\sqrt{4+9}}=2$
Squaring both sides, we get-
$$
\Rightarrow 4 h^2+9 k^2+16-12 h k-24 k+16 h=52
$$
Putting $h^2$ value in above equation we get
$$
\Rightarrow 12 h k-5 k^2-56 h+24 k+84=0
$$
for getting locus replacing $h, k$ by $x, y$ respectively
$$
\Rightarrow 12 x y-5 y^2-56 x+24 y+84=0
$$
$$
\Rightarrow h^2=-k^2+10 h-12...(1)
$$
also given, $\frac{2 h-3 k+4}{\sqrt{4+9}}=2$
Squaring both sides, we get-
$$
\Rightarrow 4 h^2+9 k^2+16-12 h k-24 k+16 h=52
$$
Putting $h^2$ value in above equation we get
$$
\Rightarrow 12 h k-5 k^2-56 h+24 k+84=0
$$
for getting locus replacing $h, k$ by $x, y$ respectively
$$
\Rightarrow 12 x y-5 y^2-56 x+24 y+84=0
$$
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