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

To solve this problem, we need to find the locus of the point of intersection of tangents drawn from a point to a given circle such that these tangents subtend a specific angle, \(120^\circ\), at the center of the circle.

1. The given circle is \((x - 2)^2 + (y - 3)^2 = 16\). This circle has center \(C(2, 3)\) and radius \(r = \sqrt{16} = 4\).
2. Let \(P(x_1, y_1)\) be a point from which tangents are drawn to the circle. These tangents make a specific angle of \(120^\circ\) with each other at the center of the circle.
3. When tangents from a point outside a circle subtend a specific angle at the center of the circle: \(O\), we use the formula: 

\[\theta = 2 \cdot \arcsin \left(\frac{r}{d}\right)\]

1. where \(d\\) is the distance from the center of the circle to the point.
2. Given that \(\theta = 120^\circ\\), we have: 

\[120^\circ = 2 \cdot \arcsin \left(\frac{4}{d}\right) \] \] which simplifies to \[ \arcsin \left(\frac{4}{d}\right) = 60^\circ \] \] So, we know that \[ \sin 60^\circ = \frac{4}{d} \] \] which implies \[ \frac{\sqrt{3}}{2} = \frac{4}{d} \] \] Solving f\]

1. The distance \(d\) from the center \((2, 3)\) to the point \((x_1, y_1)\) is given by the distance formula: 

\[\sqrt{(x_1 - 2)^2 + (y_1 - 3)^2} = \frac{8}{\sqrt{3}} \] \] Squaring both sides, we have: \[ (x_1 - 2)^2 + (y_1 - 3)^2 = \left(\frac{8}{\sqrt{3}}\right)^2 \] \] Simplifies to: \[ (x_1 - 2)^2 + (y_1 - 3)^2 = \frac{64}{3}\]

1. Expanding the terms, we have: 

\[(x_1^2 - 4x_1 + 4) + (y_1^2 - 6y_1 + 9) = \frac{64}{3} \]\]

1. Combining terms and simplifying: 

\[x_1^2 + y_1^2 - 4x_1 - 6y_1 + 13 = \frac{64}{3} \] \] \[ 3(x_1^2 + y_1^2) - 12x_1 - 18y_1 + 39 = 64 \] \] \[ 3x_1^2 + 3y_1^2 - 12x_1 - 18y_1 - 25 = 0\]

1. which matches option \(3x^2 + 3y^2 - 12x - 18y - 25 = 0\).