Let the abscissae of the two points P and Q on a circle be the roots of x²-4 x-6=0 and the ordinates of P and Q be the roots of y²+2 y-7=0. If PQ is a diameter of the circle x²+y²+2 a x+2 b y+c=0, then the value of (a+b-c) is _____________.

Solution

Maths sir · 4 steps

Write the circle in diameter form, and Vieta fills in every sum and product for you.

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. Expanding it gives only sums and products of the roots.

   x² - (x₁ + x₂)x + x₁x₂ + y² - (y₁ + y₂)y + y₁y₂ = 0

2. Read off the roots. From x² - 4x - 6 = 0: x₁ + x₂ = 4 and x₁x₂ = -6. From y² + 2y - 7 = 0: y₁ + y₂ = -2 and y₁y₂ = -7. Which x goes with which y does not matter, because only sums and products appear.
3. Substitute. The circle is x² + y² - 4x + 2y - 13 = 0. Matching it with x² + y² + 2ax + 2by + c = 0 gives 2a = -4, 2b = 2 and c = -13, so a = -2, b = 1, c = -13.
4. So. a + b - c = -2 + 1 + 13 = 12, which is option (a).

The idea to keep. Endpoints of a diameter given as roots of quadratics? Write (x-x₁)(x-x₂)+(y-y₁)(y-y₂)=0 and let Vieta's sums and products fill it in.

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