JEE Mathematics Circle MCQs Set 08

Answer: (b) x² + y² = 1

Explanation:
1. The points (1, 0) and (−1, 0) are ends of a diameter on the x-axis, so the centre lies on the perpendicular bisector x = 0.
2. Take the centre (0, k): 1 + k² = (1 − k)², so k = 0.
3. Centre (0, 0) and radius 1, so x² + y² = 1.

Teacher's Note:
1. The centre lies on the perpendicular bisector of each chord.
2. Use the centre (0, k) and equate the distances to (1, 0) and (0, 1).

Answer: (a) x² + y² = a(x + y)

Explanation:
1. The vertices are A(0, 0), B(a, 0), C(a, a) and D(0, a). The diagonal AC is a diameter.
2. The circle on the diameter joining (0, 0) and (a, a) is (x − 0)(x − a) + (y − 0)(y − a) = 0.
3. x² − ax + y² − ay = 0, i.e. x² + y² = a(x + y).

Teacher's Note:
1. The circle on a diameter with ends (x₁, y₁), (x₂, y₂) is (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0.
2. The diagonal AC is a diameter because the angle ABC is a right angle.

Answer: (c) x² + y² − 3x − 4y = 0

Explanation:
1. The circle passes through (0, 0), (3, 0) and (0, 4), so it has the form x² + y² + 2gx + 2fy = 0.
2. Putting (3, 0): 9 + 6g = 0, so g = −3/2. Putting (0, 4): 16 + 8f = 0, so f = −2.
3. Equation: x² + y² − 3x − 4y = 0.

Teacher's Note:
1. A circle through the origin has no constant term.
2. The intercepts 3 and 4 give the points (3, 0) and (0, 4).

Answer: (a) (x − 2)² + (y − 2)² = 4

Explanation:
1. Touching both axes in the first quadrant gives centre (r, r) and radius r.
2. Distance from (r, r) to 3x + 4y + 11 = 0 is (3r + 4r + 11)/5 = 5, so 7r + 11 = 25 and r = 2.
3. Circle: (x − 2)² + (y − 2)² = 4.

Teacher's Note:
1. The centre of a circle touching both axes in the first quadrant is (r, r).
2. The expression 7r + 11 is positive, so drop the absolute value.

Answer: (b) x² + y² − 6x − 6√2 y + 9 = 0

Explanation:
1. Touching the x-axis at (3, 0): centre (3, k) (or (−3, k)) and radius |k|.
2. y-intercept: 2√(r² − 3²) = 6, so r² − 9 = 9 and r² = 18, k = ±3√2.
3. With centre (3, 3√2): (x − 3)² + (y − 3√2)² = 18, i.e. x² + y² − 6x − 6√2 y + 9 = 0.
4. The other options give intercepts 0, 6√3 and 2√7 on the y-axis.

Teacher's Note:
1. For a circle touching the x-axis, the radius equals the absolute y-coordinate of the centre.
2. The y-intercept of x² + y² + 2gx + 2fy + c = 0 is 2√(f² − c).

Answer: (d) (x + 3)² + (y − 4)² = 4

Explanation:
1. x + 3 = 2 sin θ and y − 4 = 2 cos θ.
2. Squaring and adding: (x + 3)² + (y − 4)² = 4(sin²θ + cos²θ) = 4.

Teacher's Note:
1. Isolate sin θ and cos θ, then use sin²θ + cos²θ = 1.

Answer: (d) 36 < p < 47

Explanation:
1. Power of (2, 5): 4 + 25 − 16 − 60 + p = p − 47 < 0, so p < 47.
2. Centre (4, 6), radius² = 52 − p. The distances of the centre from the axes are 6 (x-axis) and 4 (y-axis).
3. Not touching or cutting the axes needs radius < 4: 52 − p < 16, so p > 36.
4. Hence 36 < p < 47.

Teacher's Note:
1. Negative power means the point lies inside the circle.
2. The smaller of the two centre-to-axis distances decides the condition.

Answer: (d) 2

Explanation:
1. The given circle has centre O(0, −√2) and radius 2. The point P(1, −1) is inside it, because OP² = 1 + (√2 − 1)² = 4 − 2√2 < 4.
2. A circle through P touching the given circle must lie inside it, touching internally: if its centre is C and radius ρ, then CP = ρ and CO = 2 − ρ.
3. So CP + CO = 2, and C lies on an ellipse with foci P and O. ρ = CP ranges between (2 − OP)/2 and (2 + OP)/2.
4. r₁ + r₂ = (2 + OP)/2 + (2 − OP)/2 = 2.

Teacher's Note:
1. Since P lies inside the given circle, only internal contact is possible.
2. CP + CO is constant, so the centre moves on an ellipse; use the extreme values of CP.

Answer: (c) 4√2

Explanation:
1. Substitute y = x + 2: 3x² + 3(x + 2)² − 29x − 19(x + 2) + 56 = 0, which simplifies to 6x² − 36x + 30 = 0, i.e. x² − 6x + 5 = 0.
2. So x = 1 or 5, giving the points (1, 3) and (5, 7).
3. Length = √(16 + 16) = 4√2.

Teacher's Note:
1. Find both intersection points by substitution.
2. Then apply the distance formula.

Answer: (d) x − 7y − 45 = 0

Explanation:
1. Centre (5/2, −5/2), radius √(25/2) = 5/√2.
2. Distance of the centre from x − 7y + 5 = 0: |5/2 + 35/2 + 5|/√50 = 25/√50 = 5/√2, so the given line is a tangent.
3. A parallel tangent is x − 7y + k = 0 with |5/2 + 35/2 + k|/√50 = 5/√2, i.e. |20 + k| = 25. k = 5 is the given line, so k = −45.
4. The other tangent is x − 7y − 45 = 0.

Teacher's Note:
1. The two parallel tangents are on opposite sides of the centre, at equal distances.
2. Use the distance formula with the line x − 7y + k = 0.

Answer: (b) y = √3 x ± 4

Explanation:
1. Slope m = tan 60° = √3.
2. Tangent: y = mx ± a√(1 + m²) = √3 x ± 2·√(1 + 3) = √3 x ± 4.

Teacher's Note:
1. Use the slope form y = mx ± a√(1 + m²) with a = 2.

Answer: (c) 16x² − 65y² − 288x + 1296 = 0

Explanation:
1. Since (9, 0) is outside the circle (81 > 16), two tangents exist. Use SS₁ = T².
2. S = x² + y² − 16, S₁ = 81 − 16 = 65 and T = 9x − 16.
3. 65(x² + y² − 16) = (9x − 16)², i.e. 65x² + 65y² − 1040 = 81x² − 288x + 256.
4. So 16x² − 65y² − 288x + 1296 = 0, equivalently 16(x − 9)² = 65y².

Teacher's Note:
1. The pair of tangents from an external point is SS₁ = T².
2. Check: 16(x − 9)² − 65y² has no xy term and vanishes at (9, 0).

Answer: (c) 2 sin⁻¹(4/9)

Explanation:
1. The distance from the centre to the point is 9 and the radius is 4. Each tangent makes an angle α/2 with the line joining the centre to the point, where sin(α/2) = 4/9.
2. So the angle between the tangents is α = 2 sin⁻¹(4/9) (tan(α/2) = 4/√65).

Teacher's Note:
1. In the right triangle formed by the centre, the external point and a point of contact, the radius is opposite to the half-angle.

Answer: (a) f² + g² + 4f + 4g + 2 = 0

Explanation:
1. Tangent length² from (f, g): to the first circle f² + g² − 6; to the second f² + g² + 3f + 3g.
2. Condition: f² + g² − 6 = 4(f² + g² + 3f + 3g).
3. 3f² + 3g² + 12f + 12g + 6 = 0, i.e. f² + g² + 4f + 4g + 2 = 0.

Teacher's Note:
1. The length of the tangent from a point equals √S₁.
2. Square the given relation to avoid the square roots.

Answer: (a) y = 2x

Explanation:
1. The normal at a point of a circle passes through the centre (0, 0).
2. Slope = 2/1 = 2, so the normal is y = 2x.

Teacher's Note:
1. Normal = line through the centre and the point.

Answer: (b) x + 2y = 1

Explanation:
1. The circle has centre (1, 0). Every normal passes through the centre.
2. A normal parallel to x + 2y = 3 is x + 2y = k with 1 + 0 = k, so k = 1.
3. Normal: x + 2y = 1.

Teacher's Note:
1. Normals are lines through the centre.

Answer: (a) (x + 4)² + y² = 16

Explanation:
1. The director circle is concentric with the given circle and has radius √2 times as large.
2. Radius² = 2 × 8 = 16, centre (−4, 0).
3. Equation: (x + 4)² + y² = 16.

Teacher's Note:
1. Tangents at right angles meet on a circle of radius r√2.

Answer: (a) a(h² + k² − a²)^{3/2} / (h² + k²)

Explanation:
1. Let d = √(h² + k²) be the distance of the point from the centre and L = √(d² − a²) the tangent length.
2. Chord of contact length = 2aL/d. Its distance from the point is d − a²/d = L²/d.
3. Area = ½ × 2aL/d × L²/d = aL³/d² = a(h² + k² − a²)^{3/2}/(h² + k²).

Teacher's Note:
1. Chord of contact: 2aL/d. Distance of the point from the chord: L²/d.

Answer: (c) x + 2y + 6 = 0

Explanation:
1. Use T = S₁. S₁ = 4 + 4 − 12 − 16 + 9 = −11.
2. T = x(−2) + y(−2) + 3(x − 2) + 4(y − 2) + 9 = x + 2y − 5.
3. x + 2y − 5 = −11, i.e. x + 2y + 6 = 0.

Teacher's Note:
1. The chord with a given midpoint is T = S₁.

Answer: (a) (6, −18/5)

Explanation:
1. The common chord is the difference: 5x − 3y − 10 = 0.
2. The tangents at the ends of this chord meet at its pole with respect to x² + y² = 12. For the chord lx + my = n the pole is (12l/n, 12m/n) = (12·5/10, 12·(−3)/10).
3. Point = (6, −18/5).

Teacher's Note:
1. The chord of contact is the common chord; the required point is its pole.
2. Compare 5x − 3y = 10 with xx₁ + yy₁ = 12.

Answer: (b) 3x + 4y = 10

Explanation:
1. Circle 1: centre (1, 3), radius 1. Circle 2: centre (−3, 1), radius 3. The distance between the centres is √20 > 4, so there are four common tangents.
2. For 3x + 4y = 10: distance from (1, 3) is |3 + 12 − 10|/5 = 1, and from (−3, 1) is |−9 + 4 − 10|/5 = 3. Both equal the radii.
3. The other options fail: for 3x + 4y = 5 the distances are 2 and 2; for 4x + 3y = 0 they are 13/5 and 9/5; for y = 2 they are 1 and 1.

Teacher's Note:
1. A common tangent has distance r₁ from the first centre and r₂ from the second.
2. Test each option with the distance formula.

Answer: (a) cut each other orthogonally

Explanation:
1. For the circles 2g₁g₂ + 2f₁f₂ = c₁ + c₂ gives orthogonality. Here g₁ = −1, f₁ = −3, c₁ = −12 and g₂ = 3, f₂ = 2, c₂ = −6.
2. 2g₁g₂ + 2f₁f₂ = −6 − 12 = −18 and c₁ + c₂ = −18.
3. The condition holds, so the circles cut orthogonally.

Teacher's Note:
1. Test the orthogonality condition 2g₁g₂ + 2f₁f₂ = c₁ + c₂.

Answer: (d) 2x² + 2y² − 7x + 2y = 0

Explanation:
1. Let the circle be x² + y² + 2gx + 2fy = 0 (c = 0).
2. Orthogonal to the second circle: 2g(0) + 2f(6) = 0 + 6, so f = 1/2.
3. Orthogonal to the first: 2g(−2) + 2f(3) = 0 + 10, so −4g + 3 = 10 and g = −7/4.
4. Circle: x² + y² − (7/2)x + y = 0, i.e. 2x² + 2y² − 7x + 2y = 0.

Teacher's Note:
1. Passing through the origin means c = 0.
2. Apply 2gg′ + 2ff′ = c + c′ with each circle.

Answer: (c) (33/4, 2)

Explanation:
1. Write the circles with unit coefficients: S₁: x² + y² − 16x + 60, S₂: x² + y² − 12x + 27, S₃: x² + y² − 16x − 12y + 84.
2. S₁ − S₂ = 0: −4x + 33 = 0, so x = 33/4. S₁ − S₃ = 0: 12y − 24 = 0, so y = 2.
3. The radical centre is (33/4, 2).

Teacher's Note:
1. Equal tangent lengths means equal powers: this is the radical centre.
2. Divide the second equation by 3 before subtracting.

Answer: (b) 1/4

Explanation:
1. The radical centre is (33/4, 2).
2. The tangent length squared equals the value of S₁ there: (33/4)² + 4 − 16(33/4) + 60 = 1089/16 + 4 − 132 + 60 = 1/16.
3. The length is 1/4.

Teacher's Note:
1. Substitute the radical centre in any one circle (with unit coefficients) and take the square root.