Circles: JEE Advanced PYQ, 2007–2026

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)

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: (A)

Any circle through the intersection points of the circle S = 0 and the line L = 0 has the form S + λL = 0. Here S + λ(x + y − 1) = 0. The origin lies on it when −4 − λ = 0, so λ = −4. The circle is x² + y² − 2x − 4y − 4 − 4x − 4y + 4 = 0, which simplifies to x² + y² − 6x − 8y = 0. Its centre is (3, 4) and its radius is √(9 + 16 + 0) = 5.

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: (D)

Completing squares gives (x − 2)² + (y − 1)² = 4, so the centre is (2, 1) and r = 2. The distance from the centre to the line is |3·2 + 4·1 + 20|/5 = 30/5 = 6. Since 6 > 2, the line does not meet the circle. The nearest point of the circle lies along the perpendicular from the centre, so the shortest distance is 6 − 2 = 4.

Answer: (A)

The centre lies on the diameter x + y = 5. It is also equidistant from A and B, so it lies on the perpendicular bisector of AB. The midpoint of AB is (3, 1) and the slope of AB is 1/2, so the bisector is y − 1 = −2(x − 3), i.e. y = −2x + 7. Solving with x + y = 5 gives x = 2, y = 3, so the centre is (2, 3). Then r² = (2 − 1)² + 3² = 10. The circle is x² + y² − 4x − 6y + 3 = 0. Put x = 0: y² − 6y + 3 = 0, whose roots differ by √(36 − 12) = √24 = 2√6. Do not take the midpoint of AB as the centre: (3, 1) does not lie on x + y = 5, because AB is a chord and not a diameter.

Answer: (D)

The circle has centre (1, −3) and r² = 1 + 9 − 6 = 4. The point P is outside the circle, since d² = CP² = 16 + 16 = 32 > 4. For any line through P, PA · PB = d² − r² = 32 − 4 = 28. This equals the value S₁ obtained by substituting P into the circle's equation.

Answer: (A)

The condition PA = 2PB gives (x − 1)² + (y − 2)² = 4[(x − 4)² + (y + 1)²]. The left side is x² + y² − 2x − 4y + 5 and the right side is 4x² + 4y² − 32x + 8y + 68. Subtracting and dividing by 3 gives x² + y² − 10x + 4y + 21 = 0. So the centre is (5, −2) and r² = 25 + 4 − 21 = 8, which gives r = 2√2. As a check, the Apollonius radius is k·AB/|1 − k²| = 2(3√2)/3 = 2√2.