The centre of the circle cutting $x^{2} + y^{2} - 2 x + 4 y - 1 = 0$ orthogonally and passing through $(0, 0)$ , $(2, 0)$ is

(\frac{3}{2}, 1)

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(1, \frac{3}{4})

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(1, \frac{- 3}{4})

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(- 1, - 3)

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Solution

The correct option is A $(1, \frac{3}{4})$
Let, $S : x^{2} + y^{2} + 2 g x + 2 f y = 0$ be the equation of the circle passing through $(0, 0)$ and $(2, 0)$
The two circles are orthogonal to each other. Hence,
$2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$
$2 g (- 1) + 2 f (2) = 0 - 1$
$4 f - 2 g = - 1$ --------(1)
The point $(2, 0)$ lies on the circle. Hence,
$4 + 4 g = 0$
$g = - 1$
$\Rightarrow f = \frac{- 3}{4}$
So, centre of circle is $(- g, - f) = (1, \frac{3}{4})$
Hence, option 'B' is correct.

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