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If one end of a diameter of the circle $x^2+y^2-4 x-6 y+11$ $=0$ is $(3,4)$, then find the coordinate of the other end of the diameter.
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Verified Answer
Here, $2 \mathrm{~g}=-4 \& 2 \mathrm{f}=-6$
$$
\Rightarrow \mathrm{g}=-2 \& \mathrm{f}=-3
$$
Hence centre is $(2,3) \&$ one end is $(3,4)$. Let other end be $(a, b)$
$$
\begin{aligned}
&\therefore \frac{a+3}{2}=2 \text { and } \frac{b+4}{2}=3 \\
&\Rightarrow a=1, b=2 \\
&\therefore(a, b) \equiv(1,2)
\end{aligned}
$$
$$
\Rightarrow \mathrm{g}=-2 \& \mathrm{f}=-3
$$
Hence centre is $(2,3) \&$ one end is $(3,4)$. Let other end be $(a, b)$
$$
\begin{aligned}
&\therefore \frac{a+3}{2}=2 \text { and } \frac{b+4}{2}=3 \\
&\Rightarrow a=1, b=2 \\
&\therefore(a, b) \equiv(1,2)
\end{aligned}
$$
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