If one end of the diameter of the circle $x^{2} + y^{2} - 8 x - 4 y + c = 0$ is $(- 3, 2)$ then other co-ordinate is

(5, 3)

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(6, 2)

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(1, - 8)

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(11, 2)

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Solution

The correct option is D $(11, 2)$
Given equation of circle is
$x^{2} + y^{2} - 8 x - 4 y + c = 0$
Center is at $(4, 2)$
Now, given one end of diameter as $(- 3, 2)$
Let the other end of the diameter be $(x_{1}, y_{1})$
$\frac{x_{1} - 3}{2} = 4, \frac{y_{1} + 2}{2} = 2$
$\Rightarrow x_{1} = 11, y_{1} = 2$
So, the other end of diameter is $(11, 2)$

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