JEE Mathematics Circle MCQs Set 07

Answer: (d) −35 < m < 15

Explanation:
1. Circle: x² + y² − 4x − 8y − 5 = 0, centre (2, 4), radius √(4 + 16 + 5) = 5.
2. Two distinct points need distance of the centre from 3x − 4y − m = 0 less than 5: |6 − 16 − m|/5 < 5.
3. |m + 10| < 25, so −35 < m < 15.

Teacher's Note:
1. A line cuts a circle at two points when the centre-to-line distance is less than the radius.
2. Solve the absolute-value inequality carefully.

Answer: (d) |a| = c

Explanation:
1. First circle: centre (a/2, 0), radius |a|/2. Second: centre (0, 0), radius c.
2. Distance between centres = |a|/2. External contact needs |a|/2 = c + |a|/2, impossible for c > 0.
3. Internal contact: |a|/2 = |c − |a|/2|, which gives c = |a|.

Teacher's Note:
1. Test both external and internal contact.
2. The first circle passes through the origin, which is the centre of the second.

Answer: (b) x² + y² − x − y = 0

Explanation:
1. The smallest circle through two points has the segment joining them as diameter.
2. (x − 1)(x − 0) + (y − 0)(y − 1) = 0.
3. x² − x + y² − y = 0, i.e. x² + y² − x − y = 0.

Teacher's Note:
1. Smallest radius means the two points are diametrically opposite.
2. Use the diameter form directly.

Answer: (b) \( \frac{10}{3} \)

Explanation:
1. Centre (1, k) and radius |k| (touching the x-axis at (1, 0)).
2. Through (2, 3): (2 − 1)² + (3 − k)² = k², so 1 + 9 − 6k = 0 and k = 5/3.
3. Diameter = 2k = 10/3.

Teacher's Note:
1. The centre lies directly above or below the point of contact.
2. Radius equals the distance of the centre from the x-axis.

Answer: (b) (5, −2)

Explanation:
1. Centre (3, k), radius |k|. Through (1, −2): (1 − 3)² + (−2 − k)² = k², so 4 + 4 + 4k = 0 and k = −2.
2. Circle: (x − 3)² + (y + 2)² = 4.
3. Test (5, −2): 4 + 0 = 4, so it lies on the circle. The others do not: (−5, 2) gives 64 + 16, (2, −5) gives 1 + 9, (−2, 5) gives 25 + 49.

Teacher's Note:
1. Use the touching condition to place the centre vertically above or below (3, 0).
2. Substitute each option in the final equation.

Answer: (a) \( \frac14 \)

Explanation:
1. T has centre (0, y) and passes through the origin, so its radius is |y|.
2. External contact: distance between centres = sum of radii: √(1 + (y − 1)²) = 1 + |y|.
3. For y > 0: 1 + y² − 2y + 1 = 1 + 2y + y², so 1 = 4y and y = 1/4. For y < 0 no solution results.
4. Radius = 1/4.

Teacher's Note:
1. Square both sides and cancel y².
2. Check both signs of y.

Answer: (b) circle of radius √2

Explanation:
1. Every line of the family passes through the intersection of 2x − 3y + 4 = 0 and x − 2y + 3 = 0, which is the point (1, 2).
2. The mirror image of P(2, 3) in a line through Q(1, 2) is at the same distance from Q as P is.
3. QP = √(1 + 1) = √2, so the image lies on the circle centred at (1, 2) with radius √2.

Teacher's Note:
1. A reflection in a line through Q preserves the distance to Q.
2. Find the fixed point of the family by solving the two lines.

Answer: (d) 3

Explanation:
1. Circle 1: centre (2, 3), radius 5. Circle 2: centre (−3, −9), radius √(9 + 81 − 26) = 8.
2. Distance between centres = √(25 + 144) = 13 = 5 + 8.
3. The circles touch externally, so there are 3 common tangents.

Teacher's Note:
1. Compare the distance of centres with the sum and difference of radii.
2. External contact gives three common tangents.

Answer: (a) a parabola

Explanation:
1. Given circle: centre (4, 4), radius √(16 + 16 + 4) = 6.
2. Let the centre of the moving circle be (h, k) with radius k (k > 0). External contact: √((h − 4)² + (k − 4)²) = 6 + k.
3. Squaring: (h − 4)² + k² − 8k + 16 = 36 + 12k + k², so (h − 4)² = 20k + 20.
4. This is a parabola.

Teacher's Note:
1. The radius equals the y-coordinate of the centre because the circle touches the x-axis.
2. A relation of the form (x − p)² = c(y + d) is a parabola.

Answer: (b) 5√3

Explanation:
1. Given circle: centre (2, −3), radius √(4 + 9 + 12) = 5. A diameter is a chord of length 10 with mid-point (2, −3).
2. Distance from the centre of S to the mid-point: √(25 + 25) = √50.
3. R² = (√50)² + 5² = 75, so R = 5√3.

Teacher's Note:
1. The line from the centre of S to the chord's mid-point is perpendicular to the chord.
2. Use Pythagoras with half the chord length.

Answer: (c) \( 3\sqrt{\frac52} \)

Explanation:
1. Orthocentre H, centroid G and circumcentre C are collinear with G dividing HC in the ratio 2 : 1, so H = 3G − 2C.
2. C = (3G − H)/2 = ((9 + 3)/2, (9 − 5)/2) = (6, 2).
3. AC = √(81 + 9) = 3√10, so the radius of the circle on AC as diameter is 3√10/2 = 3√(5/2).

Teacher's Note:
1. Use Euler line: H + 2C = 3G.
2. The radius is half the length of AC.

Answer: (a) \( \frac1{\sqrt a}=\frac1{\sqrt b}+\frac1{\sqrt c} \)

Explanation:
1. Two circles of radii r₁, r₂ touching externally and both touching a line have their points of contact with the line at a distance 2√(r₁r₂) apart.
2. The smallest circle (radius a) sits between the two larger ones, so 2√(bc) = 2√(ab) + 2√(ac).
3. Divide by 2√(abc): 1/√a = 1/√c + 1/√b.

Teacher's Note:
1. The distance between the points of contact on the x-axis is √((r₁ + r₂)² − (r₁ − r₂)²) = 2√(r₁r₂).
2. The smallest circle fits in the gap between the other two.

Answer: (d) 5

Explanation:
1. Given circle: centre (−2, 3), radius √(4 + 9 + 12) = 5; (1, −1) lies on it.
2. The centre of C lies on the line from (−2, 3) through (1, −1), beyond (1, −1): (1, −1) + ρ(3, −4)/5, where ρ is the radius of C.
3. Distance to (4, 0) equals ρ: (−3 + 3ρ/5)² + (−1 − 4ρ/5)² = ρ², giving 10 − 2ρ = 0.
4. ρ = 5.

Teacher's Note:
1. Touching circles have their centres and the point of contact on one line.
2. The centre of C is at distance ρ from the contact point along the outward direction.

Answer: (d) [12, 21]

Explanation:
1. Circle 1: centre (1, 1), radius 1. Circle 2: centre (9, 1), radius √(81 + 1 − 78) = 2.
2. For the circles to lie on opposite sides, the centres give values of opposite sign: (7 − λ)(31 − λ) < 0, so 7 < λ < 31.
3. The line must also be at distance at least 1 from (1, 1) and at least 2 from (9, 1): (λ − 7)/5 ≥ 1 gives λ ≥ 12, and (31 − λ)/5 ≥ 2 gives λ ≤ 21.
4. So 12 ≤ λ ≤ 21.

Teacher's Note:
1. Opposite sides requires both separation of the centres and enough distance from each circle.
2. A line touching a circle still leaves that circle on one side, so the end-points are included.

Answer: (a) (x² + y²)³ = 4R²x²y²

Explanation:
1. The angle at O is 90°, so AB is a diameter: A(a, 0), B(0, b) with a² + b² = 4R².
2. The foot of the perpendicular from O to x/a + y/b = 1 is (x, y) = (ab²/(a² + b²), a²b/(a² + b²)) = (ab²/4R², a²b/4R²).
3. Then x² + y² = a²b²/(4R²) and xy = a³b³/(16R⁴), so (ab)³ = 16R⁴xy and (ab)² = 4R²(x² + y²).
4. Cube the second relation and square the first: 64R⁶(x² + y²)³ = (ab)⁶ = 256R⁸x²y², so (x² + y²)³ = 4R²x²y².

Teacher's Note:
1. Right angle at the origin makes AB a diameter, so a² + b² = 4R².
2. Express x and y through a and b and eliminate them.

Answer: (b) 210

Explanation:
1. Distance of the line x + y = n from the centre is n/√2, which must be less than 4: n² < 32, so n = 1, 2, 3, 4, 5.
2. Chord length squared = 4(16 − n²/2) = 64 − 2n².
3. Sum = 5(64) − 2(1 + 4 + 9 + 16 + 25) = 320 − 110 = 210.

Teacher's Note:
1. Only lines at distance less than the radius cut chords; list the admissible n first.
2. Use chord² = 4(r² − d²).

Answer: (a) x² + y² − 4x²y² = 0

Explanation:
1. Tangent at (cos θ, sin θ): x cos θ + y sin θ = 1, meeting the axes at P(1/cos θ, 0) and Q(0, 1/sin θ).
2. Mid-point (x, y) = (1/(2cos θ), 1/(2sin θ)), so cos θ = 1/(2x) and sin θ = 1/(2y).
3. Using cos²θ + sin²θ = 1: 1/(4x²) + 1/(4y²) = 1, so x² + y² = 4x²y².

Teacher's Note:
1. Write the tangent in parametric form and read the mid-point of the intercepts.
2. Eliminate θ with cos² + sin² = 1.

Answer: (a) \( y=\sqrt{1+2x},\ x\ge0 \)

Explanation:
1. Let the centre be (h, k) in the first quadrant. Touching the y-axis gives radius h.
2. External contact with the unit circle at the origin: √(h² + k²) = 1 + h.
3. Squaring: k² = 1 + 2h, so y = √(1 + 2x) with x ≥ 0.

Teacher's Note:
1. Touching the y-axis makes the radius equal to the x-coordinate of the centre.
2. Square both sides and cancel h².

Answer: (b) c² + 6c + 7 = 0

Explanation:
1. Tangent L₁ at (1/√2, 1/√2): x + y = √2, with slope −1, so m = 1 for the perpendicular line.
2. y = x + c touches (x − 3)² + y² = 1: |3 + c|/√2 = 1, so (c + 3)² = 2.
3. c² + 6c + 9 = 2, i.e. c² + 6c + 7 = 0.

Teacher's Note:
1. Perpendicular to slope −1 means slope 1.
2. Equate the centre-to-line distance with the radius 1.

Answer: (b) 36

Explanation:
1. First circle: centre (3, 0), radius 1. Second: centre (0, 4), radius √k.
2. Distance between centres = 5.
3. External contact: 1 + √k = 5, so k = 16. Internal contact: |√k − 1| = 5, so √k = 6 and k = 36.
4. The largest value is 36.

Teacher's Note:
1. Consider both external and internal contact of the two circles.
2. Compare the two resulting values of k.