A) \[31x+35y-25=0\]
B) \[31x-35y+25=0\]
C) \[35x+31y-25=0\]
D) \[35x-31y+25=0\]
Correct Answer: B
Solution :
Let\[{{S}_{1}}\equiv 2{{x}^{2}}+2{{y}^{2}}+14x-18y+15=0\] \[\Rightarrow \] \[{{S}_{1}}\equiv {{x}^{2}}+{{y}^{2}}+7x-9y+\frac{15}{2}=0\] and \[{{S}_{2}}\equiv 4{{x}^{2}}+4{{y}^{2}}-3x-y+5=0\] \[\Rightarrow \] \[{{S}_{2}}\equiv {{x}^{2}}+{{y}^{2}}-\frac{3}{4}x-\frac{1}{4}y+\frac{5}{4}=0\] We know that, the equation of radical axis of two circles \[{{S}_{1}}\] and \[{{S}_{2}}\] is given by \[{{S}_{1}}-{{S}_{2}}=0\] \[\Rightarrow \] \[\left( {{x}^{2}}+{{y}^{2}}+7x-9y+\frac{15}{2} \right)\] \[-\left( {{x}^{2}}+{{y}^{2}}-\frac{3}{4}x-\frac{1}{4}y+\frac{5}{4} \right)=0\] \[\Rightarrow \] \[\left( 7-\frac{3}{4} \right)x+\left( -9+\frac{1}{4} \right)y+\frac{15}{2}-\frac{5}{4}=0\] \[\Rightarrow \] \[\frac{31}{4}x-\frac{35}{4}y+\frac{25}{4}=0\] \[\Rightarrow \] \[31x-35y+25=0\]
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