JEE Mathematics Circle MCQs Set 10

Answer: (a) x² + y² − 4x − 6y + 9 = 0

Explanation:
1. Let the centre be (r, k); touching the y-axis gives radius |r|.
2. Through A: (4 − r)² + (3 − k)² = r². Through B: (2 − r)² + (5 − k)² = r². Subtracting the two: k = r + 1.
3. Substituting in the first: r² − 12r + 20 = 0, so r = 2 (k = 3) or r = 10 (k = 11).
4. r = 2 gives (x − 2)² + (y − 3)² = 4, i.e. x² + y² − 4x − 6y + 9 = 0. (The other circle is x² + y² − 20x − 22y + 121 = 0.)

Teacher's Note:
1. The centre of a circle touching the y-axis is at distance r from it.
2. Test the options by substituting A and B.

Answer: (d) (0, 3)

Explanation:
1. The angle subtended by AB at a point P of the y-axis is largest when the circle through A, B, P touches the y-axis at P.
2. The circles through A and B touching the y-axis touch it at (0, 3) and (0, 11).
3. At P = (0, 3): vectors PA = (4, 0) and PB = (2, 2) give angle 45°. At P = (0, 11): PA = (4, −8), PB = (2, −6) give cos θ = 56/(√80·√40) ≈ 0.99, about 8°.
4. The largest angle (45°) is at P(0, 3).

Teacher's Note:
1. Largest angle subtended by a segment on a line occurs where a circle through its ends touches the line.

Answer: (a) a² > 2b²

Explanation:
1. The circle is x² + y² − ax − (b/2)y = 0. Let the mid-point of a chord be (s, 0). The chord is T = S₁: sx − a(x + s)/2 − (b/4)y = s² − as.
2. It passes through (a, b/2): as − a(a + s)/2 − b²/8 = s² − as, which simplifies to s² − (3a/2)s + a²/2 + b²/8 = 0.
3. Two chords need two distinct real values of s, so the discriminant is positive: 9a²/4 − 2a² − b²/2 > 0, i.e. a²/4 > b²/2.
4. Hence a² > 2b².

Teacher's Note:
1. The mid-point of the chord is (s, 0): use T = S₁.

Answer: (a) (3 − 2√2)t² − 2√2 t + 2 = 0

Explanation:
1. Let the centre be (1 − t, k): touching x = 1 from the left gives radius t = 1 − (x-coordinate). (A circle through the origin cannot lie to the right of x = 1.)
2. Distance of the centre from x + y = 2 equals t: |1 − t + k − 2|/√2 = t, so k = 1 + t ± √2 t.
3. The circle passes through the origin: (1 − t)² + k² = t², so 1 − 2t + k² = 0.
4. With k = 1 + (1 ± √2)t: 1 − 2t + (1 + (1 ± √2)t)² = 0, i.e. (3 ± 2√2)t² + 2(1 ± √2)t − 2t + 2 = 0, which simplifies to (3 ± 2√2)t² ± 2√2 t + 2 = 0.
5. The upper signs give (3 + 2√2)t² + 2√2 t + 2 = 0, which has no real root (negative discriminant). The lower signs give (3 − 2√2)t² − 2√2 t + 2 = 0, with roots about 0.74 and 15.74, both valid radii.

Teacher's Note:
1. Write the centre as (1 − t, k) and use the two distance conditions.
2. Check the roots 0.74 and 15.74 numerically.

Answer: (a) 4√5 + 18 cot⁻¹(2/√5)

Explanation:
1. Let f(x) = x² − 4x − α². One root in (−1, 0) needs f(−1) > 0 and f(0) < 0: 5 − α² > 0 and α ≠ 0, so |α| < √5.
2. So (a, α) lies in the disc x² + y² < 9 with |α| < √5: the disc without the two caps cut off by the lines α = ±√5.
3. Each cap has half-angle θ where cos θ = √5/3 (so tan θ = 2/√5); its area is 9θ − 9 sin θ cos θ = 9θ − 2√5.
4. Required area = 9π − 2(9θ − 2√5) = 9π − 18θ + 4√5. Since π/2 − θ = cot⁻¹(2/√5), 9π − 18θ = 18 cot⁻¹(2/√5).
5. Area = 4√5 + 18 cot⁻¹(2/√5), about 24.08.

Teacher's Note:
1. Find the range of α from the sign conditions at −1 and 0.
2. The region is a disc minus two circular segments.

Answer: (b) x² + y² + gx + fy + c/2 = 0

Explanation:
1. Let the chord be lx + my = 1. The lines joining the origin to its ends are given by x² + y² + (2gx + 2fy)(lx + my) + c(lx + my)² = 0.
2. A right angle at the origin means the sum of the coefficients of x² and y² is zero: 2 + 2gl + 2fm + c(l² + m²) = 0.
3. The foot of the perpendicular from the origin is (x₁, y₁) = (l, m)/(l² + m²), so l = x₁/ρ, m = y₁/ρ with ρ = x₁² + y₁² and l² + m² = 1/ρ.
4. Then 2ρ + 2gx₁ + 2fy₁ + c = 0: the locus is x² + y² + gx + fy + c/2 = 0.

Teacher's Note:
1. Homogenise S with the chord and use the condition for perpendicular lines.

Answer: (b) (4√3 − 3)x − (4 + 3√3)y − (39 − 2√3) = 0

Explanation:
1. The circle has centre C(1, 2) and radius 5. The radius vector CP = (−3, −4); for anticlockwise motion the direction of the tangent at P is (4, −3).
2. After reflection the path passes through C, so the image C′ of C in the mirror lies on the line through P along (4, −3): C′ = P + λ(4, −3)/5.
3. PC = 5, PC′ = λ and PC ⟂ PC′ (tangent), so CC′² = 25 + λ². The mirror is the perpendicular bisector of CC′, at distance |25 − λ²|/(2√(25 + λ²)) from P. Setting this equal to 5/2: (25 − λ²)² = 25(25 + λ²), so λ² = 75 and λ = 5√3.
4. C′ = (−2 + 4√3, −2 − 3√3). The perpendicular bisector of CC′ (direction (4√3 − 3, −4 − 3√3), mid-point ((4√3 − 1)/2, −3√3/2)) is (4√3 − 3)x − (4 + 3√3)y − (39 − 2√3) = 0.

Teacher's Note:
1. Use the image of the centre in the mirror: it lies on the incident ray.
2. The mirror is the perpendicular bisector of the segment joining the centre and its image.

Answer: (b) x² + y² − 10x − 4y + 4 = 0

Explanation:
1. Let the centre be (h, k). Distance from each line is 3: 5h + 12k − 10 = ±39 and 5h − 12k − 40 = ±39.
2. Adding the two: 10h − 50 = ±39 ± 39, so h = 5 (or 12.8, −2.8). h = 5 with 5h + 12k − 10 = 39 gives k = 2 and 5h − 12k − 40 = −39 ✓; the other choices give a centre outside the first quadrant.
3. C₁ has centre (5, 2) and radius 3. For C₂ a chord of length 8 is at distance 3 from the centre, so R² = 3² + 4² = 25.
4. C₂: (x − 5)² + (y − 2)² = 25, i.e. x² + y² − 10x − 4y + 4 = 0.

Teacher's Note:
1. The centre is at distance 3 (the radius) from both lines.
2. A chord of length 8 at distance 3 gives R² = 9 + 16.

Answer: (c) G.P.

Explanation:
1. For the point (x₁, y₁) with x₁² + y₁² = a², the chord of contact is xx₁ + yy₁ = b².
2. Its distance from the origin is b²/√(x₁² + y₁²) = b²/a.
3. Touching x² + y² = c² means this distance equals c: c = b²/a, so b² = ac.
4. a, b, c are in G.P.

Teacher's Note:
1. The distance of the chord xx₁ + yy₁ = b² from the origin is b²/a.

Answer: (d) 2x² + 2y² − 2px − 2qy + p² + q² − a² = 0

Explanation:
1. Let M(h, k) be the mid-point of a chord. The circle on the chord as diameter passes through (p, q) because the angle there is 90°, so M is at distance equal to the half-chord from (p, q).
2. Half-chord² = a² − (h² + k²) (since OM ⟂ chord). So (h − p)² + (k − q)² = a² − h² − k².
3. 2h² + 2k² − 2ph − 2qk + p² + q² − a² = 0.

Teacher's Note:
1. The mid-point of the hypotenuse of a right triangle is equidistant from all three vertices.

Answer: (d) x² + y² − 2dx + d² − b = 0

Explanation:
1. For a circle with centre (h, k) and radius R, the line ℓx + my + 1 = 0 is a tangent if (ℓh + mk + 1)² = R²(ℓ² + m²).
2. Expanding: ℓ²(h² − R²) + m²(k² − R²) + 2hk ℓm + 2hℓ + 2km + 1 = 0.
3. Compare with aℓ² − bm² + 2dℓ + 1 = 0: hk = 0, k = 0, h = d, h² − R² = a and k² − R² = −b. Then R² = d² − a = b (consistent as a + b = d²).
4. Circle: (x − d)² + y² = b, i.e. x² + y² − 2dx + d² − b = 0.

Teacher's Note:
1. Write the tangency condition and compare coefficients of ℓ², m², ℓm, ℓ, m.

Answer: (b) (−4, 4) and (−1/2, 1/2)

Explanation:
1. Let S: x² + y² + 2gx + 2fy + c = 0. Orthogonality with x² + y² − 4 = 0: 0 = c − 4, so c = 4.
2. The centre (−g, −f) lies on 2x − 2y + 9 = 0: −2g + 2f + 9 = 0, so f = g − 9/2.
3. S: x² + y² + 4 − 9y + 2g(x + y) = 0 for all g. The fixed points satisfy x + y = 0 and x² + y² − 9y + 4 = 0.
4. With y = −x: 2x² + 9x + 4 = 0, so x = −4 or −1/2. The points are (−4, 4) and (−1/2, 1/2).

Teacher's Note:
1. Write the family in the form S₁ + gL = 0; the fixed points are the common solutions of S₁ = 0 and L = 0.

Answer: (d) c² = a²(2 + m²)

Explanation:
1. The centre is on the x-axis: (h, 0), with radius² = h² + a².
2. Tangency to mx − y + c = 0: (mh + c)² = (h² + a²)(1 + m²), i.e. h² − 2mch + a²(1 + m²) − c² = 0. The two circles correspond to the roots h₁, h₂.
3. Orthogonality: (h₁ − h₂)² = r₁² + r₂² = h₁² + h₂² + 2a², so −2h₁h₂ = 2a² and h₁h₂ = −a².
4. The product of the roots is a²(1 + m²) − c² = −a², so c² = a²(2 + m²).

Teacher's Note:
1. Circles through (0, ± a) have centres on the x-axis.
2. Use the product of roots of the quadratic in h.

Answer: (c) 9x + 6y = 65

Explanation:
1. cos(tan⁻¹(2/3)) = 3/√13 and sin = 2/√13. So A′ = (3, 2) and B′ = (6, 4).
2. C₁: (x − 3)² + (y − 2)² = 52/9. C₂: (x − 6)² + (y − 4)² = 13/9.
3. Subtracting: 6x + 4y − 39 − 39/9 = 0, i.e. 6x + 4y = 130/3, which is 9x + 6y = 65.

Teacher's Note:
1. Find A′ and B′ by rotation, then subtract the two circle equations.

Answer: (d) a² − 4ab + b² = 0

Explanation:
1. The circles have centres (r, r) and radius r. Through P: (a − r)² + (b − r)² = r², i.e. r² − 2(a + b)r + a² + b² = 0. Let the roots be r₁, r₂.
2. The distance between centres is √2|r₁ − r₂|. Orthogonality: 2(r₁ − r₂)² = r₁² + r₂², so r₁² + r₂² = 4r₁r₂ and (r₁ + r₂)² = 6r₁r₂.
3. Using r₁ + r₂ = 2(a + b) and r₁r₂ = a² + b²: 4(a + b)² = 6(a² + b²), so 2(a + b)² = 3(a² + b²).
4. This gives a² − 4ab + b² = 0.

Teacher's Note:
1. Use the sum and product of the radii from the quadratic.

Answer: (a) x² + y² − 4x − 6y − 4 = 0

Explanation:
1. The common chord of S and a circle Cᵢ is a diameter of Cᵢ, so the centre of Cᵢ lies on the radical axis: the power of the centre cᵢ with respect to S equals −rᵢ².
2. Let S: x² + y² + 2gx + 2fy + c = 0. Circle 1: centre (0, 0), r² = 4: c = −4.
3. Circle 2: centre (3, 4), r² = 15: 25 + 6g + 8f − 4 = −15, so 3g + 4f = −18. Circle 3: centre (−1, 2), r² = 1 + 4 + 2 = 7: 5 − 2g + 4f − 4 = −7, so −g + 2f = −4.
4. Solving: f = −3 and g = −2. S: x² + y² − 4x − 6y − 4 = 0.

Teacher's Note:
1. The centre of the bisected circle lies on the common chord (radical axis).

Answer: (d) 2

Explanation:
1. BD = √(64 + 36) = 10 (the diagonal).
2. In triangle ABD, the incircle touches BD at E with BE = (AB + BD − AD)/2 = (8 + 10 − 6)/2 = 6.
3. In triangle CBD, BF = (BC + BD − CD)/2 = (6 + 10 − 8)/2 = 4.
4. EF = BE − BF = 2.

Teacher's Note:
1. The tangent length from a vertex to the incircle is (sum of the two sides at the vertex − opposite side)/2.

Answer: (a) obtuse-angled

Explanation:
1. Take the common tangent as the x-axis. The points of contact with the tangent are 2√(r₁r₂) apart for S₁, S₂, 2√(r₁r) for S₁, S and 2√(r r₂) for S, S₂.
2. Since S lies between them: 2√(r₁r) + 2√(r r₂) = 2√(r₁r₂), so √r = √(r₁r₂)/(√r₁ + √r₂).
3. Example r₁ = 4, r₂ = 1: r = 4/9, centres S₁(0, 4), S₂(4, 1), S(8/3, 4/9). Vectors from S: (−8/3, 32/9) and (4/3, 5/9) have dot product −32/9 + 160/81 < 0, so the angle at S is obtuse.
4. In general, with u = √r₁, v = √r₂ and r = u²v²/(u + v)², the dot product of the vectors from S to the other centres works out to r(r − 2uv). Since r < uv/4, it is negative for every r₁, r₂, so the angle at S is always obtuse.

Teacher's Note:
1. Compute with a concrete case such as r₁ = 4, r₂ = 1 and find the angle at S.

Answer: (a) (km/(m² + n²), kn/(m² + n²))

Explanation:
1. A circle through O meeting the axes at P(p, 0) and Q(0, q) is x² + y² − px − qy = 0, with mp + nq = k.
2. Substitute q = (k − mp)/n: x² + y² − ky/n + p(−x + my/n) = 0 for every p.
3. The fixed point satisfies x = my/n and x² + y² − ky/n = 0: y²(m²/n² + 1) = ky/n, so y = kn/(m² + n²) and x = km/(m² + n²).

Teacher's Note:
1. Write the family as S + pL = 0 and solve S = 0, L = 0.

Answer: (a) 2(x + y) − a = 2xy/a

Explanation:
1. The triangle is right-angled at the origin, so its circumcentre is the mid-point P(h, k) of the hypotenuse, with vertices (2h, 0) and (0, 2k).
2. The hypotenuse is kx + hy − 2hk = 0. The circle has centre (a, a) and radius a, so its distance from the line is a: |a(h + k) − 2hk| = a√(h² + k²).
3. Squaring: a²(h + k)² − 4ahk(h + k) + 4h²k² = a²(h² + k²), so 2a²hk − 4ahk(h + k) + 4h²k² = 0.
4. Dividing by 2hk: a² − 2a(h + k) + 2hk = 0, i.e. 2(h + k) − a = 2hk/a.

Teacher's Note:
1. The circumcentre of a right triangle is the mid-point of the hypotenuse.

Answer: (c) x² + y² + 2x − 3y + 2 + √7(x + 2y − 2) = 0

Explanation:
1. The line AB is x + 2y − 2 = 0 (check: A gives 0 + 2 − 2 = 0 and B gives −2 + 4 − 2 = 0). The circle on AB as diameter is (x − 0)(x + 2) + (y − 1)(y − 2) = 0, i.e. x² + y² + 2x − 3y + 2 = 0.
2. Every circle through A and B is x² + y² + 2x − 3y + 2 + λ(x + 2y − 2) = 0. Its centre is (−(2 + λ)/2, (3 − 2λ)/2).
3. Radius² = (2 + λ)²/4 + (3 − 2λ)²/4 − (2 − 2λ) = 5(λ² + 1)/4.
4. Setting this equal to 10: λ² = 7, so λ = ±√7. The circle with λ = √7 is the option; λ = √5 and √3 give radius² 7.5 and 5; λ = 7 gives 62.5.

Teacher's Note:
1. All circles through A and B belong to the family S + λL = 0 with L the line AB.

Answer: (b) (a − b)/h = (a′ − b′)/h′

Explanation:
1. Every curve through the four points is S + λS′ = 0. The four points are concyclic if this family contains a circle.
2. A circle needs equal coefficients of x² and y² and no xy term: a + λa′ = b + λb′ and h + λh′ = 0.
3. From the second, λ = −h/h′. Substituting: a − b = −λ(a′ − b′) = (h/h′)(a′ − b′).
4. So (a − b)/h = (a′ − b′)/h′.

Teacher's Note:
1. A circle has no xy term and equal coefficients of x² and y².