x2 + y2 – 3x + y = 0
x2 + y2 + x + 3y = 0
x2 + y2 + 2y – 1 = 0
x2 + y2 + x + y = 0
Given that the mirror image of the orthocenter lies on the circumcircle:
The image of the point (1, 1) reflected over the x-axis is (1, -1). The image of the point (1, 1) reflected over the line \(x+y+1\)=0 is (-2, -2).
Therefore, the circle passing through both (1, -1) and (-2, -2) is determined.
Thus, the circle represented by the equation x2 + y2 + x + 3y = 0 satisfies this condition.
Hence the correct option is B
List - I | List - II | ||
(P) | α equals | (1) | (-2, 4) |
(Q) | r equals | (2) | \(\sqrt5\) |
(R) | A1 equals | (3) | (-2, 6) |
(S) | B1 equals | (4) | 5 |
(5) | (2, 4) |