It is given that in ∆ABC,∠A=90∘, so equation of circumcircle of ∆ABC, where B(2,−4) and C(1,5) ∵B and C are end points of diameter of the circumcircle of ∆ABC, so equation of circumcircle is (x−2)(x−1)+(y+4)(y−5)=0 ⇒‌‌x2+y2−3x−y−18=0