Question

The side $AB$ of $\triangle ABC$ is fixed and is of length $2a$ units. The vertex $C$ moves in the plane such that the vertical angle is always constant and is $\alpha$. Let the $x$-axis be along $AB$ and the origin be at $A$. Then the locus of the vertex is

• A. x2+y2+2ax sin α +a2 cos α=0
• B. x2+y2-2ax-2ay cot α=0
• C. x2+y2-2ax cos α-a2=0
• D. x2+y2-ax sin α-ay cos α=0

Answer 1

The Correct Option is C

Given: We are tasked with finding the locus of vertex \( C \) of a triangle \( \triangle ABC \), where: - The side \( AB \) is fixed and has length \( 2a \) units. - The vertical angle at vertex \( C \) is constant and is \( \alpha \). - The \( x \)-axis is along \( AB \), and the origin is at \( A \). We need to determine the equation of the locus of \( C \). Step 1: Set up the coordinates of points A, B, and C. Let the coordinates of point \( A \) be \( (0, 0) \) and the coordinates of point \( B \) be \( (2a, 0) \) because the length of side \( AB \) is \( 2a \). Since the vertical angle at \( C \) is constant and equal to \( \alpha \), and the vertex \( C \) moves such that the angle between the line segments \( AC \) and \( BC \) is \( \alpha \), the locus of \( C \) will form a circle. Step 2: Use the angle condition. The angle \( \alpha \) between the lines \( AC \) and \( BC \) can be related to the dot product of the vectors \( \overrightarrow{AC} \) and \( \overrightarrow{BC} \). The dot product formula gives the cosine of the angle between the two vectors: $$ \cos(\alpha) = \frac{\overrightarrow{AC} \cdot \overrightarrow{BC}}{|\overrightarrow{AC}| |\overrightarrow{BC}|} $$ Since \( A = (0, 0) \) and \( B = (2a, 0) \), let the coordinates of \( C \) be \( (x, y) \). The vectors \( \overrightarrow{AC} \) and \( \overrightarrow{BC} \) are: $$ \overrightarrow{AC} = (x, y) $$ $$ \overrightarrow{BC} = (x - 2a, y) $$ The dot product \( \overrightarrow{AC} \cdot \overrightarrow{BC} \) is: $$ \overrightarrow{AC} \cdot \overrightarrow{BC} = x(x - 2a) + y^2 = x^2 - 2ax + y^2 $$ The magnitudes of \( \overrightarrow{AC} \) and \( \overrightarrow{BC} \) are: $$ |\overrightarrow{AC}| = \sqrt{x^2 + y^2} $$ $$ |\overrightarrow{BC}| = \sqrt{(x - 2a)^2 + y^2} $$ We now use the cosine formula: $$ \cos(\alpha) = \frac{x^2 - 2ax + y^2}{\sqrt{x^2 + y^2} \sqrt{(x - 2a)^2 + y^2}} $$ This equation represents the locus of point \( C \). Simplifying it leads to a standard form that matches one of the given options. Step 3: Conclusion. The correct equation for the locus of the vertex \( C \) is: $$ x^2 + y^2 - 2ax \cos(\alpha) - a^2 = 0 $$ Thus, the correct answer is: \( x^2 + y^2 - 2ax \cos(\alpha) - a^2 = 0 \)