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Find the equation of the circle whose diameter is the common chord of the circles x^{2}+y^{2}+4 x+3 y+2=0 and x^{2}+y^{2}+6 x+3 y+2=0

Option: 1

x^2+ y^2+3y+2=0


Option: 2

x^2+y^2=1


Option: 3

x^2+ y^2+3y+3=0


Option: 4

None of these


Answers (1)

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\\x^{2}+y^{2}+4 x+3 y+2=0\\ x^{2}+y^{2}+6 x+3 y+2=0\\ \text{Equation of common chord of two circles is }\\ S-S'=0\Rightarrow x=0\\\\ \text{To get points of intersection of circles, put x=0 in equation of any circle }\\ y^2+3y+2=0 \Rightarrow y=-1 ,-2\\ \text{Points of intersection } (0,-1),(0,-2) \\ \text{These are the end points of diameter of required circle. So using diameter form, the required circle is} \\ (x-0)(x-0)+(y+1)(y+2)=0 \\x^2+ y^2+3y+2=0
 

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