If (0,3 \/ 4) is the radical centre of the circles S \equiv x^{2}+y^{2}+\alpha x+6 y=0, S^{'} \equiv x^{2}+y^{2}+2 \alpha x+\alpha y+6=0 and S^{' '} \equiv x^{2}+y^{2}+6 \alpha x-\alpha y+3=0 then the distance between the radical centre and the centre of the circle S^{'}=0 is

TS EAMCET 18-July-2022 Shift 2 Solved Paper

Section: Mathematics

Question : 55 of 160

Marks: +1, -0

If (0,3∕4) is the radical centre of the circles S≡x2+y2+αx+6y=0, S′≡x2+y2+2αx+αy+6=0 and S′′≡x2+y2+6αx−αy+3=0 then the distance between the radical centre and the centre of the circle S′=0 is

8
15
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√65
4
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√5
4

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