Question

The locus of the point of intersection of tangents 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:

For tangents from an external point: \begin{itemize} \item Angle between tangents \(= 2\alpha \Rightarrow OP = \dfrac{r}{\sin\alpha}\) \item Locus problems often result in a circle \item Always identify centre and radius first \end{itemize}

Answer 1

The Correct Option is A

Step 1: The given circle has centre \[ C(2,3) \] and radius \[ r = 4. \]
Step 2: If the angle between tangents drawn from a point \(P\) to a circle is \(2\alpha\), then: \[ CP = \frac{r}{\sin \alpha}. \] Here, \[ 2\alpha = 120^\circ \Rightarrow \alpha = 60^\circ. \]
Step 3: Distance of point \(P\) from the centre: \[ CP = \frac{4}{\sin 60^\circ} = \frac{4}{\frac{\sqrt{3}}{2}} = \frac{8}{\sqrt{3}}. \]
Step 4: Hence, the locus of point \(P\) is a circle with centre \((2,3)\) and radius \( \frac{8}{\sqrt{3}} \). \[ (x - 2)^2 + (y - 3)^2 = \frac{64}{3} \] Multiplying by 3: \[ 3x^2 + 3y^2 - 12x - 18y - 25 = 0 \]