Circles: JEE Main PYQ, 2020–2025

Answer: (A)

The circle meets the x-axis where y = 0, so x² − 6x + 8 = 0, which factors as (x − 2)(x − 4) = 0. The points are (2, 0) and (4, 0), so the intercept length is 2.

Answer: (C)

The first circle has centre (1, 2) and r₁² = 1 + 4 − 1 = 4, so r₁ = 2. The second has centre (4, 6) and r₂² = 16 + 36 − k = 52 − k. The distance between the centres is √(9 + 16) = 5. External contact requires r₁ + r₂ = 5, so r₂ = 3. Then 52 − k = 9 and k = 43.

Answer: (A)

For the ellipse a² = 25 and b² = 9, so c² = 16 and the foci are (±4, 0). The circle with F₁F₂ as diameter is x² + y² = 16. Put y² = 9(1 − x²/25) into it: x² + 9 − 9x²/25 = 16, so 16x²/25 = 7 and x² = 175/16. Then y² = 16 − 175/16 = 81/16. The four points are (±5√7/4, ±9/4), the vertices of a rectangle with sides 2|x| = 5√7/2 and 2|y| = 9/2. The area is (5√7/2)(9/2) = 45√7/4.

Answer: (A)

The circle is (x + 2)(x − 6) + y² = 0, which is x² + y² − 4x − 12 = 0. The centre is (2, 0) and the radius is 4. Substituting (2, 4): 4 + 16 − 8 − 12 = 0, so it lies on the circle. The other points give nonzero values.

Answer: (A)

First make the coefficients of x² and y² equal to 1 by dividing by 4. This gives x² + y² − 3x + 4y − 11/4 = 0, so g = −3/2 and f = 2. The centre is (−g, −f) = (3/2, −2). Its distance from the origin is √(9/4 + 4) = √(25/4) = 5/2.

Answer: (A)

The power of P with respect to the circle is 5² − 9 = 16, so PA·PB = 16. With PB = 2PA, 2PA² = 16, so PA = 2√2 and PB = 4√2. Then AB = PB − PA = 2√2. The distance d of the centre from the line satisfies d² = 9 − (√2)² = 7. The line is y = m(x − 5), so d = 5m/√(1 + m²). Setting 25m² = 7(1 + m²) gives 18m² = 7, so m = √14/6.

Answer: (D)

If a circle touches a line, the radius equals the perpendicular distance from the centre to that line. The distance from (−5, 2) to the y-axis (x = 0) is |−5| = 5, so the radius is 5.

Answer: (A)

(x + 2)² + (y − 5)² = 4 + 25 − c. For radius 4, 29 − c = 16, so c = 13.

Answer: (A)

The circle has centre C(−1, 2) and radius r = 3. Triangles PAC and PBC are congruent right triangles with the right angle at the point of contact. Each has area ½·r·PA, so the quadrilateral has area r·PA, where PA = √(PC² − r²). To minimise the area, minimise PC. PC is least when P is the foot of the perpendicular from C to the line. The distance from C to the line is |−3 + 8 − 40|/5 = 7, which exceeds r = 3, so P lies outside the circle and the tangents exist. Then PA² = 49 − 9 = 40, so PA = 2√10. The minimum area is 3 × 2√10 = 6√10.

Answer: (A)

Complete the square: (x − 1)² + (y − 2)² = 4, so the centre is (1, 2) and the radius is 2. Tangency requires the distance from the centre to the line to equal the radius: |3(1) − 4(2) + k| / √(9 + 16) = 2. This gives |k − 5| = 10, so k = 15 or k = −5. The largest value is 15.