Step 1: Convert the circle equation to standard form
The given circle equation is:
\[
x^2 + y^2 - 4x - 6y - 12 = 0
\]
Group the terms:
\[
(x^2 - 4x) + (y^2 - 6y) = 12
\]
Complete the square:
\[
(x^2 - 4x + 4) + (y^2 - 6y + 9) = 12 + 4 + 9
\Rightarrow (x - 2)^2 + (y - 3)^2 = 25
\]
So, the circle has center at \( (2, 3) \) and radius \( R = \sqrt{25} = 5 \)
Step 2: Parametric form of a point on a circle
For a circle with center \( (h, k) \) and radius \( R \), the parametric coordinates of a point are:
\[
x = h + R \cos \theta,\quad y = k + R \sin \theta
\]
Substitute \( h = 2 \), \( k = 3 \), \( R = 5 \)
For point \( P\left( \frac{\pi}{3} \right) \):
\[
x_P = 2 + 5 \cos\left( \frac{\pi}{3} \right) = 2 + 5 \cdot \frac{1}{2} = 4.5
\]
\[
y_P = 3 + 5 \sin\left( \frac{\pi}{3} \right) = 3 + 5 \cdot \frac{\sqrt{3}}{2} = 3 + \frac{5\sqrt{3}}{2}
\]
For point \( Q\left( \frac{2\pi}{3} \right) \):
\[
x_Q = 2 + 5 \cos\left( \frac{2\pi}{3} \right) = 2 + 5 \cdot \left( -\frac{1}{2} \right) = -0.5
\]
\[
y_Q = 3 + 5 \sin\left( \frac{2\pi}{3} \right) = 3 + 5 \cdot \frac{\sqrt{3}}{2} = 3 + \frac{5\sqrt{3}}{2}
\]
Step 3: Use distance formula to find chord length PQ
Note that both points have the same \( y \)-coordinate, so PQ is a horizontal chord:
\[
PQ = |x_P - x_Q| = |4.5 - (-0.5)| = 5
\]
Final Answer:
5