Let the parabola y=x²+p x-3, meet the coordinate axes at the points P, Q and R . If the circle C with centre at (-1,-1) passes through the points P, Q and R, then the area of △ P Q R is :

Solution

Maths sir · 4 steps

The centre sits above the midpoint of chord PQ, so the two x-intercepts must average to -1.

1. Name the points. On the y-axis put x = 0: R = (0, -3). On the x-axis, P and Q are (x₁, 0) and (x₂, 0), where x₁, x₂ are the roots of x² + px - 3 = 0. So x₁ + x₂ = -p.
2. Use the centre. PQ is a chord of circle C, and the perpendicular from the centre bisects any chord. PQ lies along the x-axis, so the centre's x-coordinate is the midpoint of x₁ and x₂.

   (x₁ + x₂)/2 = -1 ⇒ -p = -2 ⇒ p = 2

3. Find P and Q. x² + 2x - 3 = (x + 3)(x - 1) = 0 gives P = (1, 0) and Q = (-3, 0). Check: the distances from (-1, -1) to P and to R are both √5, so the circle fits.
4. So. Base PQ = 1 - (-3) = 4 lies on the x-axis, and the height is the distance of R from that axis, which is 3. Area = 1/2 × 4 × 3 = 6, option (b).

The idea to keep. When a circle passes through two points on an axis, the centre lies on the perpendicular bisector of that chord. Match the centre's coordinate to the midpoint, then use the sum of the roots.

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