Key Concepts and Formulas

• Equation of a Circle: $(x-h)^2 + (y-k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.
• Tangent Properties: A tangent to a circle is perpendicular to the radius at the point of tangency. Tangents from an external point to a circle have equal lengths.
• Area of a Triangle: Area $= \frac{1}{2} \times \text{base} \times \text{height}$ or Area $= \frac{1}{2}ab\sin C$.

Step-by-Step Solution

1. Find the Center and Radius of the Circle

The equation of the circle is given as $x^2 + y^2 - 4x + 3 = 0$. We complete the square to rewrite the equation in standard form to determine the center and radius.

• Explanation: Converting to standard form $(x-h)^2 + (y-k)^2 = r^2$ makes identifying the center $(h,k)$ and radius $r$ straightforward.
• Working: $x^2 - 4x + y^2 + 3 = 0$ $(x^2 - 4x + 4) + y^2 + 3 - 4 = 0$ $(x - 2)^2 + y^2 - 1 = 0$ $(x - 2)^2 + y^2 = 1$
• Result: The center of the circle is $C(2, 0)$ and the radius is $r = \sqrt{1} = 1$.

2. Visualize the Geometry and Determine OA

Tangents $OA$ and $OB$ are drawn from the origin $O(0,0)$ to the circle. Since $CA$ is a radius and $OA$ is a tangent at $A$, $\angle OAC = 90^\circ$. Similarly, $\angle OBC = 90^\circ$. We can use the Pythagorean theorem in $\triangle OAC$ to find the length of $OA$.

• Explanation: $OA$ and $OB$ are tangents from the origin to the circle. The radius is perpendicular to the tangent at the point of contact.
• Working: In right triangle $\triangle OAC$, we have $OC^2 = OA^2 + AC^2$. The coordinates of $O$ are $(0,0)$ and the coordinates of $C$ are $(2,0)$. Therefore, $OC = \sqrt{(2-0)^2 + (0-0)^2} = \sqrt{4} = 2$. Also, $AC = r = 1$. So, $2^2 = OA^2 + 1^2$, which means $4 = OA^2 + 1$, and $OA^2 = 3$. Thus, $OA = \sqrt{3}$. Since $OA$ and $OB$ are tangents from the same point, $OA = OB = \sqrt{3}$.

3. Find the Angle AOB

We can find $\angle AOC$ using trigonometry in $\triangle OAC$.

• Explanation: We need the angle between the two tangents to calculate the area of the triangle.
• Working: $\sin(\angle AOC) = \frac{AC}{OC} = \frac{1}{2}$ Therefore, $\angle AOC = \frac{\pi}{6} = 30^\circ$. Since $\angle AOC = \angle BOC$, we have $\angle AOB = 2 \times \angle AOC = 2 \times 30^\circ = 60^\circ$.

4. Calculate the Area of Triangle OAB

We know $OA = OB = \sqrt{3}$ and $\angle AOB = 60^\circ$. We can use the formula for the area of a triangle given two sides and the included angle.

• Explanation: We have two sides and the included angle; hence we use the formula $\frac{1}{2}ab\sin C$.
• Working: Area of $\triangle OAB = \frac{1}{2} \times OA \times OB \times \sin(\angle AOB)$ Area of $\triangle OAB = \frac{1}{2} \times \sqrt{3} \times \sqrt{3} \times \sin(60^\circ)$ Area of $\triangle OAB = \frac{1}{2} \times 3 \times \frac{\sqrt{3}}{2}$ Area of $\triangle OAB = \frac{3\sqrt{3}}{4}$

Common Mistakes & Tips

• Completing the square errors: Double-check your calculations when completing the square to avoid errors in the center and radius.
• Incorrect trigonometric ratios: Make sure you use the correct trigonometric ratios when finding the angles.
• Forgetting the formula for the area of a triangle: Remember the area formula when two sides and the included angle are known.

Summary

We first found the center and radius of the circle by completing the square. Then, using the properties of tangents and the Pythagorean theorem, we determined the length of the tangent segment from the origin to the circle. Next, we calculated the angle between the tangents. Finally, we used the two sides and included angle formula to compute the area of the triangle formed by the origin and the points of tangency.

Final Answer

The final answer is $\boxed{\frac{3 \sqrt{3}}{4}}$, which corresponds to option (B).