A) \[15x-12y=0\]
B) \[3x-2y=12\]
C) \[5x-4y=48\]
D) None of these
Correct Answer: C
Solution :
The radical axis of the circle \[{{S}_{1}}=0\] and \[{{S}_{2}}=0\] is \[{{S}_{1}}-{{S}_{2}}=0\] \[\therefore \ ({{x}^{2}}+{{y}^{2}}-144)-({{x}^{2}}+{{y}^{2}}-15x+12y)=0\] \[\Rightarrow 15x-12y-144=0\Rightarrow 5x-4y=48\].You need to login to perform this action.
You will be redirected in
3 sec