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**Hint:**We first try to form the given circle in its general form of $ {{\left( x-\alpha \right)}^{2}}+{{\left( y-\beta \right)}^{2}}={{r}^{2}} $ to find the centre and the radius. Then we try to find the circumcircle of the $ \Delta OPQ $ . We prove that it’s a cyclic quadrilateral. We find the arc creator as the diameter of the new circle. We find the midpoint of that diameter to find the solution to the problem.

**Complete step by step answer:**

It’s given that O is the origin and OP and OQ are the tangents from the origin to the circle $ {{x}^{2}}+{{y}^{2}}-6x+4y+8=0 $ . We transform it as $ {{\left( x-3 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=5 $ . O is the origin.

Equating with the general equation of circle $ {{\left( x-\alpha \right)}^{2}}+{{\left( y-\beta \right)}^{2}}={{r}^{2}} $ , we get the centre as $ C\equiv \left( 3,-2 \right) $ and the radius as $ \sqrt{5} $ units.

PQ is the chord of contact. So, OPCQ becomes cyclic quadrilateral.

We can see CP and CQ are radii. OP and OQ are tangents which means $ \angle OPC=\angle OQC={{90}^{\circ }} $ . The circumcircle of $ \Delta OPQ $ will intersect point C also.

Basically, the segment OC becomes the arc creator of the circumcircle as it creates the right angle at points P and Q. We can define the OC segment as the diameter of the circle.

The middle point of the points O and C becomes the circumcentre of $ \Delta OPQ $ .

The points are $ O\equiv \left( 0,0 \right) $ and $ C\equiv \left( 3,-2 \right) $ .

The middle point which is circumcentre of $ \Delta OPQ $ will be \[\left( \dfrac{3+0}{2},\dfrac{-2+0}{2} \right)\equiv \left( \dfrac{3}{2},-1 \right)\].

**The correct option is B**.

**Note:**

We also can use the general form to find the circumcentre where the centre is always \[\left( \dfrac{-g}{2},\dfrac{-f}{2} \right)\] for the circle $ {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 $ .We can sue the formula for only cyclic quadrilaterals.

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