A) \[x-3y-15=0\]
B) \[x+3y-15=0\]
C) \[x-3y+15=0\]
D) \[2x+3y-16=0\]
Correct Answer: B
Solution :
The points (1, 3) and (2, 6) \[{{(x-1)}^{2}}+{{(y-3)}^{2}}=0\] \[\Rightarrow \] \[{{x}^{2}}+1-2x+{{y}^{2}}+9-6y=0\] \[{{x}^{2}}+{{y}^{2}}-2x-6y+10=0\] ?(i) \[{{(x-2)}^{2}}+(y-6)=0\] \[\Rightarrow \] \[{{x}^{2}}+4-4x+{{y}^{2}}+36-12y=0\] \[{{x}^{2}}+{{y}^{2}}-4x-12y+36=0\] ?(ii) Therefore radical axis of the circles is given by equation (i) equation (ii)\[=0\] \[2x+6y-30=0\]or\[x+3y=-15=0\]
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