Question

If the tangents at the points P and Q on the circle x² + y² – 2x + y = 5 meet at the point R  \(( \frac{9}{4},2 ) \)then the area of the triangle PQR is 

• A. \(\frac{13}{8}\)
• B. \(\frac{13}{4}\)
• C. \(\frac{5}{8}\)
• D. \(\frac{5}{4}\)

Answer 1

The Correct Option is C

To solve this problem, we need to find the area of the triangle PQR where P and Q are points on the circle, and the tangents at P and Q intersect at point R \(\left( \frac{9}{4}, 2 \right)\).

The equation of the circle is \(x^2 + y^2 - 2x + y = 5\). 

1. First, rewrite the circle's equation in the standard form \((x - a)^2 + (y - b)^2 = r^2\).
   • Complete the square for x: \(x^2 - 2x = (x - 1)^2 - 1\).
   • Complete the square for y: \(y^2 + y = (y + \frac{1}{2})^2 - \frac{1}{4}\).
   • Thus, the circle's equation becomes:
     \((x - 1)^2 + (y + \frac{1}{2})^2 = 6.25\).
2. The equation of the tangent to a circle \((x - a)^2 + (y - b)^2 = r^2\) at a point \((x_1, y_1)\) is:
   • \((x_1 - a)(x - a) + (y_1 - b)(y - b) = r^2\)
3. Since the tangents at P and Q meet at R \(( \frac{9}{4}, 2 )\) and knowing the center of the circle, we use the power of the point theorem which states:
   • The power of point R is the same as the power of point R with respect to the tangents.
   • Power of R with respect to the circle: \((\frac{9}{4} - 1)^2 + (2 + \frac{1}{2})^2 - 6.25\)
   • \((\frac{5}{4})^2 + (\frac{5}{2})^2 - 6.25\)
   • \(= \frac{25}{16} + \frac{25}{4} - 6.25\)
   • \(= \frac{25}{16} + \frac{100}{16} - \frac{100}{16} = \frac{25}{16}\)
4. Now, using the property of tangents from an external point, the length of the tangent is the square root of the power of the point:
   • Length = \(\sqrt{\frac{25}{16}} = \frac{5}{4}\)
5. The area of triangle PQR is half the product of the lengths of the tangents RP and RQ:
   • Since the tangents are equal, the area is \(\frac{1}{2} \times RP \times RQ = \frac{1}{2} \times \left(\frac{5}{4}\right)^2\)
   • \(= \frac{1}{2} \times \frac{25}{16} = \frac{25}{32}\)

Therefore, the correct answer is \(\frac{5}{8}\).