First MethodAs we know that, mirror image of orthocenter Image of (1,1), in X-axis is (1,−1). and Image of (1,1) in x+y+1=0 is (− ∴ The required circle will be passing th (−2,−2).∴ The required circle will be passing through both (1,−1) and (−2,−2) Onlyx2+y2+x+3y=0 satisfy both.Second MethodOne of the vertex is the point of intersection of X-axis and the line x+y+1=0, say A, which is A≡(−1,0).Let vertex B(λ,−λ−1).Since, AC⊥BH∴λ=1⇒B=(1,−2)Let vertex C be (μ,0).∵AH⊥BC∴( Slope of AH)×( Slope of BC)=−1​21​×μ−12​=−1⇒μ=0Now, centroid of △ABC is (0,3−2​)As we know that, centroid (G) divides the line joining circumcentre (O) and orthocentre (H) in the ratio 1:2.2α+1=0 and 2β+1=−2⇒α=−1/2 and β=−3/2⇒ Circumcentre =(2−1​,2−3​)​∴ Equation of circumcircle which is passing through C(0,0) is x2+y2+x+3y=0is x2+y2+x+3y=0Third MethodEquation of circle passing through C(0,0) isx2+y2+2gx+2fy=0Then,1−2g=0⇒g=1/25+1−4f=0⇒f=3/2​∴ Equation of circumcircle is x2+y2+2×21​x+2×23​y=0x2+y2+x+3y=0