Question

The Cartesian coordinates of a point P with parameter \(\theta =\frac{5\pi }{6}\) on the circle \(x^{2}+y^{2}-2x+4y+1=0\) are...

• A. \((1,0)\)
• B. \((0,-2+\sqrt{3})\)
• C. \((-1,-2)\)
• D. \((-\sqrt{3}+1,-1)\)

Hint:

Find centre and radius, then use x = h + r cos(theta) and y = k + r sin(theta).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Rewrite the circle as \((x-1)^2+(y+2)^2=4\), so the centre is \((1,-2)\) and radius \(r=2\).

Step 2: Parametric Form
\(x=1+2\cos\theta,\quad y=-2+2\sin\theta\).

Step 3: Substitute theta
For \(\theta=\dfrac{5\pi}{6}\): \(\cos\theta=-\dfrac{\sqrt{3}}{2}\) and \(\sin\theta=\dfrac{1}{2}\).
\(x=1-\sqrt{3}\), \(y=-2+1=-1\).

Final Answer:
The point is \((1-\sqrt{3},-1)\). \[ \boxed{\text{(D) }(-\sqrt{3}+1,-1)} \]