The locus of point of intersection of tangent drawn to the circle (x – 2)² + (y – 3)² = 16, which sub-stends an angle of 120° is

The correct option is:
3x² + 3y² – 12x – 18y – 25 = 0

Step-by-Step Explanation:

Given the equation of the circle:
${(x-2)}^2+{(y-3)}^2=16$

From this equation, we can identify:
1. The center of the circle, $C=(2,3)$.
2. The radius of the circle, $R=\sqrt{16}=4$.

Let $P(h,k)$ be the point of intersection of the tangents drawn to the circle. The angle subtended by the tangents at $P$ is $120°$.

Let $T$ be a point of tangency on the circle. The line joining the center $C$ to the point of intersection $P$ bisects the angle between the tangents. Therefore, the angle $\angle TPC$ is:
$\angle TPC=\frac{120°}{2}=60°$

In the right-angled triangle $CTP$ (where $\angle CTP=90°$ because the radius is perpendicular to the tangent at the point of contact):
$sin(60°)=\frac{\text{Opposite}}{\text{Hypotenuse}}=\frac{CT}{CP}$

Here, $CT=R=4$ and $CP$ is the distance between $C(2,3)$ and $P(h,k)$:
$sin(60°)=\frac{4}{CP}$

Since $sin(60°)=\frac{\sqrt{3}}{2}$, we have:
$\frac{\sqrt{3}}{2}=\frac{4}{CP}$

Rearranging the equation to find $CP$:
$CP=\frac{8}{\sqrt{3}}$

Squaring both sides:
$C{P}^2=\frac{64}{3}$

Using the distance formula for $C{P}^2$:
${(h-2)}^2+{(k-3)}^2=\frac{64}{3}$

Expanding the terms:
$(h^2-4h+4)+(k^2-6k+9)=\frac{64}{3}$
$h^2+k^2-4h-6k+13=\frac{64}{3}$

Multiply the entire equation by 3 to eliminate the fraction:
$3h^2+3k^2-12h-18k+39=64$

Subtracting 64 from both sides:
$3h^2+3k^2-12h-18k-25=0$

Replacing $(h,k)$ with $(x,y)$ to get the equation of the locus:
$3x^2+3y^2-12x-18y-25=0$