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If the radical axis of two circles is 2x+3y=1 and the equation of one of these circles is x^{2}+y^{2}+4 x+6 y+9=0, then find the equation of the other circle is

Option: 1

x^{2}+y^{2}+2 x+3 y+8=0 or x^{2}+y^{2}+6x+9y+12=0


Option: 2

x^{2}+y^{2}+2 x+3 y+9=0 or x^{2}+y^{2}+6x+9y+9=0


Option: 3

x^{2}+y^{2}+2 x+3 y+10=0 or x^{2}+y^{2}+6x+9y+8=0


Option: 4

None of these


Answers (1)

\\\text{Given equation of radical axis } 2x+3y=1 \\ \text{Given equation of circle } x^{2}+y^{2}+4 x+6 y+9=0\\ \therefore 2x+3y-1=(x^{2}+y^{2}+4 x+6 y+9)-S_2\\ S_2=x^{2}+y^{2}+2 x+3 y+10\\ \text{Hence the other circle is }x^{2}+y^{2}+2 x+3 y+10=0

We can have another circle satisfying the same criteria. To get that circle 

\\2x+3y-1=S_2-(x^{2}+y^{2}+4 x+6 y+9) \\S_2=x^{2}+y^{2}+6x+9y+8 \\\text{The other circle is }x^{2}+y^{2}+6x+9y+8=0

 

 

 

Posted by

Ramraj Saini

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