Question

If \(P\left( \frac{\pi}{4} \right)\), \(Q\left( \frac{\pi}{3} \right)\) are two points on the circle \(x^2 + y^2 - 2x - 2y - 1 = 0\), then the slope of the tangent to this circle which is parallel to the chord \(PQ\) is:

• A. \(2 + \sqrt{2} - \sqrt{3} - \sqrt{6}\)
• B. \(2 + \sqrt{2} + \sqrt{3} + \sqrt{6}\)
• C. \(\sqrt{2} - \sqrt{3}\)
• D. \(2 + \sqrt{2}\)

Hint:

For problems involving tangents and chords to a circle, use the properties of the circle and the relationship between the center and the tangent.
- Utilize trigonometric identities for angular points on a circle to compute slopes of lines related to the points.

Answer 1

The Correct Option is A

Step 1: Equation of the circle.
The given equation of the circle is: \[ x^2 + y^2 - 2x - 2y - 1 = 0. \] We complete the square for both \(x\) and \(y\) terms to write the equation in standard form: \[ (x-1)^2 + (y-1)^2 = 3. \] Thus, the center of the circle is \((1, 1)\) and the radius is \(\sqrt{3}\). Step 2: Equation of the chord \(PQ\).
The points \(P\left( \frac{\pi}{4} \right)\) and \(Q\left( \frac{\pi}{3} \right)\) lie on the circle, and the slope of the chord \(PQ\) is the difference in their \(y\)-coordinates divided by the difference in their \(x\)-coordinates. Hence, the slope of the chord \(PQ\) is: \[ \text{slope of } PQ = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\sin \left( \frac{\pi}{3} \right) - \sin \left( \frac{\pi}{4} \right)}{\cos \left( \frac{\pi}{3} \right) - \cos \left( \frac{\pi}{4} \right)}. \] After substituting the trigonometric values for the angles \(\frac{\pi}{4}\) and \(\frac{\pi}{3}\), we simplify to find the slope of the tangent that is parallel to this chord. \[ \boxed{2 + \sqrt{2} - \sqrt{3} - \sqrt{6}}. \]