⇒3AP=AQ Let A=(x,y) then 3AP=AQ⇒9AP2=AQ2 ⇒9(x−1)2+9y2=(x+1)2+y2 ⇒9x2−18x+9+9y2=x2+2x+1+y2 ⇒8x2−20x+8y2+8=0 ⇒x2+y2−
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x+1=0 ∴ A lies on the circle given by eq (1). As B and C also follow the same condition, they must lie on the same circle. ∴ Centre of circumcircle of ∆ABC = Centre of circle given by (1)=(