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**Hint:**Let \[P(h,k)\] be the pole of the circle \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\], then the equation of the polar line from the point is \[hx+ky+g\left( x+h \right)+f\left( y+k \right)+c=0\]. We can find the equation of the polar by substituting the coordinates of the pole in the above equation. It should be noted that the coefficient of the square terms in the equation of the circle is one, so if it is not one, make it by dividing by the coefficient.

**Complete step by step answer:**

Let the coordinates of the pole be \[\left( h,k \right)\]. The equation of the circle is \[2{{x}^{2}}+2{{y}^{2}}-3x+5y-7=0\], dividing both sides of the equation by 2, we get \[{{x}^{2}}+{{y}^{2}}-\dfrac{3}{2}x+\dfrac{5}{2}y-\dfrac{7}{2}=0\].

The equation of polar for the point having coordinates \[\left( h,k \right)\] is \[hx+ky-\dfrac{3}{4}\left( x+h \right)+\dfrac{5}{4}\left( y+k \right)-\dfrac{7}{2}=0\]. Simplifying the equation, it can be written as \[\left( 4h-3 \right)x+\left( 4k+5 \right)y+\left( -3h+5k-14 \right)\].

We are given that the equation of the polar is \[9x+y-28=0\]. So the above equation and this equation represent the same line. Comparing the coefficients of the line, we get

\[\dfrac{\left( 4h-3 \right)}{9}=\dfrac{\left( 4k+5 \right)}{1}=\dfrac{\left( -3h+5k-14 \right)}{-28}\]

Taking the first and third expressions, we get \[\dfrac{\left( 4h-3 \right)}{9}=\dfrac{\left( -3h+5k-14 \right)}{-28}\]. Taking the second and third and expression, we get \[\dfrac{\left( 4k+5 \right)}{1}=\dfrac{\left( -3h+5k-14 \right)}{-28}\].

By solving the above two equations, we get \[h=3\And k=-1\]. Hence the coordinates of the pole are \[\left( 3,-1 \right)\].

We can also show the pole and polar on the graph as follows,

**Hence, the correct option is C.**

**Note:**Pole and polar are one of the important parts of the circle, many questions can be based on these parts. It should be remembered that while writing the equation of polar for a pole, the coefficients of square terms in the equation of the circle is one. If the equation of the circle does not have coefficients as one, then make it one before solving the question.

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