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Find centre and radius of the circle 2x2+ 2y+ 8x- 6y+ 7 = 0.
1. (-2, - \(\frac{3}{2}\) )  and \(\sqrt{\frac{11}{2}}\) units. 
2. ( 2,  \(\frac{3}{2}\) ) and \(\sqrt{\frac{11}{2}}\)units.
3. (-2, - \(\frac{3}{2}\) ) and \(\frac{\sqrt{11}}{2}\) units.
4. (-2, \(\frac{3}{2}\) ) and \(\frac{\sqrt{11}}{2}\) units. 

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Correct Answer - Option 4 : (-2, \(\frac{3}{2}\) ) and \(\frac{\sqrt{11}}{2}\) units. 

Concept: 

General form of the equation of a circle, x2 + y2 + 2gx + 2fy + c = 0

Centre is (-g, -f) or  \(\rm \left ( \frac{-\ coefficient \ of \ x}{2},\ \frac{-\ coefficient \ of \ y}{2} \right )\) , where g , f and c are constant. 

Radius = \(\rm\sqrt{g^{2}+f^{2}-c}\)  

 

Calculation: 

Given equation of circle is 2x2+ 2y+ 8x- 6y+ 7 = 0  

⇒ x+ y+ 4x - 3y + \(\frac{7}{2}\) = 0             ....(i)

On compare eq. (i)  with standard equation of circle x2 + y2 + 2gx + 2fy + c = 0 

We get  ,  g = 2 , f = \(\frac{-3}{2}\) and c = \(\frac{7}{2}\) 

As we know that centre of circle is (-g, -f)

centre (-2, \(\frac{3}{2}\) ) . 

And radius of circle = \(\rm\sqrt{g^{2}+f^{2}-c}\) 

⇒ radius = \(\sqrt{2^{2}+\left ( \frac{-3}{2} \right )^{2}-\frac{7}{2}}\) = \(\sqrt{4+\frac{9}{4}-\frac{7}{2}}\) 

radius = \(\frac{\sqrt{11}}{2}\) units. 

The correct option is 4. 

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