A) \[\left( \frac{5}{3},0 \right)\]
B) (0,0)
C) \[\left( \frac{1}{3},0 \right)\]
D) (3,0)
Correct Answer: A
Solution :
Let\[P(1,0)\]and\[Q(-1,0),A(x,y)\] Given:\[\frac{AP}{AQ}=\frac{BP}{BQ}=\frac{CP}{CQ}=\frac{1}{2}\] \[\Rightarrow \]\[2AP=AQ\]\[\Rightarrow \]\[4{{(AP)}^{2}}=A{{Q}^{2}}\] \[\Rightarrow \]\[4[{{(x-1)}^{2}}+{{y}^{2}}]={{(x+1)}^{2}}+{{y}^{2}}\] \[\Rightarrow \]\[4({{x}^{2}}+1-2x)+4{{y}^{2}}={{x}^{2}}+1+2x+{{y}^{2}}\] \[\Rightarrow \]\[3{{x}^{2}}+3{{y}^{2}}-8x-2x+4-1=0\] \[\Rightarrow \]\[3{{x}^{2}}+3{{y}^{2}}-10x+3=0\] \[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}-\frac{10}{3}x+1=0\] ?(1) \[\therefore \]A lies on the circle given by (1). As B and C also follow the same condition. \[\therefore \]Centre of circumcircle of \[\Delta ABC=\]centre of circle given by (1)\[=\left( \frac{5}{3},0 \right)\]
You need to login to perform this action.
You will be redirected in
3 sec
