Let the abscissae of the two points P and Q be the roots of 2x² - rx + p = 0 and the ordinates of P and Q be the roots of x² - sx - q = 0. If the equation of the circle described on PQ as diameter is 2(x² + y²) - 11x - 14y - 22 = 0, then 2r + s - 2q + p is equal to __________.

Solution

Maths sir · 4 steps

Write the diameter form with Vieta's sums and products, then compare it with the given circle.

1. Diameter form. If P(x₁, y₁) and Q(x₂, y₂) are the ends of a diameter, the circle is (x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0. Expanded, it uses only the sums and products of the coordinates.
2. Use the roots. From 2x² - rx + p = 0: x₁ + x₂ = r/2 and x₁ x₂ = p/2. From x² - sx - q = 0: y₁ + y₂ = s and y₁ y₂ = -q.

   x² + y² - r/2x - sy + p/2 - q = 0

3. Compare coefficients. Divide the given circle by 2 so that x² has coefficient 1: x² + y² - 11/2x - 7y - 11 = 0. This gives r = 11, s = 7 and p/2 - q = -11, that is p - 2q = -22.
4. So. 2r + s - 2q + p = 22 + 7 + (p - 2q) = 29 - 22 = 7. You never need p and q separately, only the combination p - 2q.

The common slip: 14. Comparing coefficients without first dividing the given equation by 2 gives r = 22, s = 14, p - 2q = -44, and the total 14 instead of 7.

The idea to keep. For a circle on a chord as diameter, you never need the actual endpoints. Sums and products of the roots fill in the equation, and you compare it with the given circle once x² has coefficient 1.

Answer from the NTA final answer key. Solution by PrepWiser.