Consider a triangle Δ whose two sides lie on the x-axis and the line x+y+1=0. If the orthocenter of Δ is (1, 1), then the equation of the circle passing through the vertices of the triangle Δ is
(A)x2+y2−3x+y=0
(B)x2+y2+x+3y=0✓ Correct
(C)x2+y2+2y−1=0
(D)x2+y2+x+y=0
Answer: (B)
Solution
One vertex is the intersection of the x-axis and x+y+1=0, giving A(-1,0). Let vertex B be (α,-α-1) on the line x+y+1=0. Since line AC ⊥ BH, we get α=1, so B(1,-2). Let vertex C be (β,0) on the x-axis. Since line AH ⊥ BC, we have m_AH · m_BC = -1, which gives (1/2)·(2/(β-1)) = -1, so β=0. The centroid of ΔABC is (0,-2/3). The centroid G divides the line joining the circumcenter O and orthocenter H in the ratio 1:2. Using the section formula with G(0,-2/3) and H(1,1), we find the circumcenter is (-1/2,-3/2). The equation of the circumcircle passing through C(0,0) is x2+y2+x+3y=0.
JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Single correct option
