Question

The equation of the radical axis of the circles $ 2{{x}^{2}}+2{{y}^{2}}+14x-18y+15=0 $ and $ 4{{x}^{2}}+4{{y}^{2}}-3x-y+5=0 $ , is

• A. $ 31x+35y-25=0 $
• B. $ 31x-35y+25=0 $
• C. $ 35x+31y-25=0 $
• D. $ 35x-31y+25=0 $

Answer 1

Answer: (B)

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 $