The pole of the straight line $9 x + y - 28 = 0$ with respect to the circle ${2 x}^{2} + {2 y}^{2} - 3 x + 5 y - 7 = 0$ , is

(3, 1)

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(1, 3)

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(3, - 1)

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(- 3, 1)

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Solution

The correct option is C $(3, - 1)$
let $(h, k)$ be the pole $9 x + y - 28$ with respect to ${2 x}^{2} + {2 y}^{2} - 3 x + 5 y - 7 = 0$
then equation of polar $T = 0$
$2 h x + 2 k y - 3 (\frac{h + x}{2}) + 5 (\frac{k + y}{2}) - 7 = 0$
$\Rightarrow (2 h - \frac{3}{2}) x + (2 k + \frac{5}{2}) y + \frac{5 k - 3 h - 14}{2} = 0$
On comparing with $9 x + y - 28 = 0$
we get $\frac{(2 h - \frac{3}{2})}{9} = (2 k + \frac{5}{2}) = \frac{\frac{5 k - 3 h - 14}{2}}{- 28}$
Solving for, $h$ & $k$ we get $(h, k) = (3, - 1)$
Ans: C

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