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If the coordinates at one end of a diameter of the circle $x^{2}+y^{2}-8 x-4 y+c=0$ are (-3,2) , then the coordinates at the other end are
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The correct answer is:
(11,2)
The centre of the given circle is $\mathrm{C} \equiv(4,2)$
Let $A \equiv(-3,2)$
If $(\alpha, \beta)$ are the coordinates of the other end of the diameter, then, as the middle ploint of the diameter is the centre,
$\therefore \quad \frac{\alpha-3}{2}=4$ and $\frac{\beta+2}{2}=2 \Rightarrow \alpha=11, \beta=2$
Thus, the coordinates of the other end of diameter are (11,2)
Let $A \equiv(-3,2)$
If $(\alpha, \beta)$ are the coordinates of the other end of the diameter, then, as the middle ploint of the diameter is the centre,
$\therefore \quad \frac{\alpha-3}{2}=4$ and $\frac{\beta+2}{2}=2 \Rightarrow \alpha=11, \beta=2$
Thus, the coordinates of the other end of diameter are (11,2)
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