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If the coordinates of one end of a diameter of the circle \(x^2+y^2+4 x-8 y+5=0\), is \((2,1)\), the coordinates of the other end is
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Verified Answer
The correct answer is:
\((-6,7)\)
Hints: \(x^2+y^2+9 x-8 y+5=0\)
Centre circle \((-2,4)\)
\(\begin{aligned}
& \frac{\mathrm{h}+2}{2}=-2 \\
& \mathrm{~h}=-4-2=-6 \\
& \frac{\mathrm{k}+1}{2}=4 \Rightarrow \mathrm{k}=7 \\
& (\mathrm{~h}, \mathrm{k}) \rightarrow(-6,7)
\end{aligned}\)
Centre circle \((-2,4)\)
\(\begin{aligned}
& \frac{\mathrm{h}+2}{2}=-2 \\
& \mathrm{~h}=-4-2=-6 \\
& \frac{\mathrm{k}+1}{2}=4 \Rightarrow \mathrm{k}=7 \\
& (\mathrm{~h}, \mathrm{k}) \rightarrow(-6,7)
\end{aligned}\)
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