A) \[x+y+1=0\]
B) \[x-y+1=0\]
C) \[x+y-8=0\]
D) \[x-y-8=0\]
Correct Answer: C
Solution :
Let the radius of both circles are ‘r’. Now, equation of circle with centre at \[(2,3)\] is \[{{S}_{1}}\equiv {{(x-2)}^{2}}+{{(y-3)}^{2}}={{r}^{2}}\] …(i) and equation of circle with centre at \[(5,6)\] is \[{{S}_{2}}\equiv {{(x-5)}^{2}}+{{(y-6)}^{2}}={{r}^{2}}\] ….(ii) Now, the equation common chord \[\equiv \] Radical axis of \[{{S}_{1}}\] and \[{{S}_{2}}=0\] \[\equiv ({{S}_{1}}-{{S}_{2}})=0\] \[\equiv [{{(x-2)}^{2}}]+[{{(y-3)}^{2}}]\] \[-{{[(x-5)]}^{2}}-{{[(y-6)]}^{2}}=0\] \[\equiv {{x}^{2}}+{{y}^{2}}+4-4x+9-6x\] \[-{{x}^{2}}-{{y}^{2}}-25-36+10x+12y=0\] \[\equiv 6x+6y-48=0\] Common chord \[\equiv x+y-8=0\]
You need to login to perform this action.
You will be redirected in
3 sec
