Question

The locus of point of intersection of tangent drawn to the circle \( (x - 2)^2 + (y - 3)^2 = 16 \), which subtends an angle of \( 120^\circ \) is

• A. \( 3x^2 + 3y^2 - 12x - 18y - 25 = 0 \)
• B. \( x^2 + y^2 - 12x - 18y - 25 = 0 \)
• C. \( 3x^2 + 3y^2 + 12x + 18y - 25 = 0 \)
• D. \( x^2 + y^2 + 12x + 18y - 25 = 0 \)

Hint:

When solving problems involving the locus of intersection of tangents, use the geometry of the circle and the known properties of the tangents to derive the equation.

Answer 1

The Correct Option is A

Step 1: Understand the problem.
The equation \( (x - 2)^2 + (y - 3)^2 = 16 \) represents a circle with center \( (2, 3) \) and radius \( 4 \). The problem asks for the locus of the point of intersection of tangents that subtend an angle of \( 120^\circ \). Step 2: Use the property of the tangents.
The angle between the tangents drawn from a point outside the circle is given by the formula: \[ \cos\theta = \frac{r}{d} \] where \( r \) is the radius of the circle and \( d \) is the distance from the center of the circle to the point of intersection. For \( \theta = 120^\circ \), we have: \[ \cos(120^\circ) = -\frac{1}{2} \] Substitute into the formula: \[ -\frac{1}{2} = \frac{4}{d} \quad \Rightarrow \quad d = 8 \] Step 3: Derive the equation of the locus.
The distance from the center of the circle to the point of intersection is \( 8 \), and by using the equation of the tangent, we derive the required locus: \[ 3x^2 + 3y^2 - 12x - 18y - 25 = 0 \]