Question

The equation of circumcircle of a ∆ABC is $x^2 + y^2 + 3x + y - 6 = 0$ if $A = (1, -2)$, $B = (-3, 2)$ and the vertex C varies, then the locus of orthocentre of ∆ABC is a

• A. Circle
• B. Parabola
• C. Ellipse
• D. Straight Line

Answer 1

Answer: (A)

Circle

Answer 2

The circumcenter O of the given circle $x^2 + y^2 + 3x + y - 6 = 0$ is found by comparing with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$. Here, $2g = 3 \implies g = 3/2$ and $2f = 1 \implies f = 1/2$. Thus, the circumcenter is $O = (-g, -f) = (-3/2, -1/2)$.

The position vector of the orthocenter H of a triangle ABC is related to the position vectors of its vertices A, B, C and the circumcenter O by the relation: $\vec{h} = \vec{a} + \vec{b} + \vec{c} - 2\vec{o}$

Let $A = (x_A, y_A) = (1, -2)$, $B = (x_B, y_B) = (-3, 2)$, $C = (x_C, y_C)$, and $H = (x_H, y_H)$. The coordinates of the orthocenter are: $x_H = x_A + x_B + x_C - 2x_O = 1 + (-3) + x_C - 2(-3/2) = 1 - 3 + x_C + 3 = x_C + 1$ $y_H = y_A + y_B + y_C - 2y_O = -2 + 2 + y_C - 2(-1/2) = 0 + y_C + 1 = y_C + 1$

From these equations, we can express the coordinates of C in terms of the coordinates of H: $x_C = x_H - 1$ $y_C = y_H - 1$

Since vertex C lies on the circumcircle, its coordinates must satisfy the circumcircle's equation: $x_C^2 + y_C^2 + 3x_C + y_C - 6 = 0$

Substitute the expressions for $x_C$ and $y_C$ in terms of $x_H$ and $y_H$: $(x_H - 1)^2 + (y_H - 1)^2 + 3(x_H - 1) + (y_H - 1) - 6 = 0$

Expand and simplify the equation: $(x_H^2 - 2x_H + 1) + (y_H^2 - 2y_H + 1) + (3x_H - 3) + (y_H - 1) - 6 = 0$ $x_H^2 - 2x_H + 1 + y_H^2 - 2y_H + 1 + 3x_H - 3 + y_H - 1 - 6 = 0$

Combine like terms: $x_H^2 + y_H^2 + (-2x_H + 3x_H) + (-2y_H + y_H) + (1 + 1 - 3 - 1 - 6) = 0$ $x_H^2 + y_H^2 + x_H - y_H - 8 = 0$

Replacing $(x_H, y_H)$ with the general coordinates $(x, y)$ to represent the locus, we get: $x^2 + y^2 + x - y - 8 = 0$

This is the standard equation of a circle. Therefore, the locus of the orthocentre of ∆ABC is a circle.