The set of values of k, for which the circle C:4x² + 4y² - 12x + 8y + k = 0 lies inside the fourth quadrant and the point ( 1, - 1/3 ) lies on or inside the circle C, is :

Solution

Maths sir · 4 steps

Divide by 4 first; the quadrant gives a lower bound on k and the point gives an upper bound.

1. Centre and radius. Dividing by 4 gives x² + y² - 3x + 2y + k/4 = 0. So the centre is (3/2, -1) and r² = 9/4 + 1 - k/4 = (13 - k)/4.
2. Stay in quadrant. The centre is 1 unit below the x-axis and 3/2 units right of the y-axis. To stay inside the fourth quadrant without touching either axis, the radius must be less than the smaller distance, so 0 < r < 1, that is 0 < r² < 1.

   0 < (13 - k)/4 < 1 ⇒ 9 < k < 13

3. Point on or inside. Substitute (1, -1/3) into the original left side and require S ≤ 0: 4 + 4/9 - 12 - 8/3 + k ≤ 0. This gives k - 92/9 ≤ 0, so k ≤ 92/9.
4. So. Both conditions hold together for 9 < k ≤ 92/9, because 92/9 ≈ 10.2 is less than 13. This is the interval (9, 92/9], option (d).

The idea to keep. A circle lies inside a quadrant when its radius is less than the distance from its centre to the nearer axis. A point is on or inside a circle when S(x₁, y₁) ≤ 0.

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