JEE Mathematics Circle MCQs Set 09

Answer: (d) (1) and (4) only

Explanation:
1. The point (3, −6) is in the fourth quadrant, so the centre is (r, −r) with radius r.
2. (3 − r)² + (−6 + r)² = r², so r² − 18r + 45 = 0 and r = 3 or 15.
3. r = 3: (x − 3)² + (y + 3)² = 9, i.e. x² + y² − 6x + 6y + 9 = 0. r = 15: x² + y² − 30x + 30y + 225 = 0.
4. Statements (1) and (4) are correct; (2) and (3) have centres in the second quadrant.

Teacher's Note:
1. A circle touching both axes in the fourth quadrant has centre (r, −r).

Answer: (b) (2) and (3) only

Explanation:
1. The centre lies on the perpendicular bisector of the two points: x − y = 5, so the centre is (k + 5, k) with radius |k|.
2. (k + 4)² + (k + 2)² = k², so k² + 12k + 20 = 0 and k = −2 or −10.
3. k = −2: centre (3, −2), radius 2: x² + y² − 6x + 4y + 9 = 0. k = −10: centre (−5, −10), radius 10: x² + y² + 10x + 20y + 25 = 0.
4. Statements (2) and (3) are correct.

Teacher's Note:
1. Touching the x-axis means radius = |y-coordinate of the centre|.

Answer: (a) (2) and (4) only

Explanation:
1. The centre lies on x = 1/2: (1/2, k), radius r = √(1/4 + k²).
2. The circle passes through the origin, which is inside x² + y² = 9, so the contact is internal: distance of centres = 3 − r, i.e. r = 3 − r and r = 3/2.
3. k² = 9/4 − 1/4 = 2, so k = ±√2.
4. The centres are (1/2, √2) and (1/2, −√2): statements (2) and (4).

Teacher's Note:
1. The distance from the origin to the centre equals the radius, because the circle passes through the origin.

Answer: (b) (1) and (4) only

Explanation:
1. The line is 4x + 3y − 12 = 0. The distance of (c, c) from it is |7c − 12|/5, which must equal c.
2. 7c − 12 = 5c gives c = 6; 7c − 12 = −5c gives c = 1.
3. Both values give circles in the first quadrant (c = 1 is the incircle of the triangle, c = 6 an excircle): statements (1) and (4).

Teacher's Note:
1. Write the line in the form ax + by + c = 0 and use the distance formula.

Answer: (b) (3) and (4) only

Explanation:
1. Solving the two lines gives the point (7, 1), which is inside the circle (50 < 100).
2. Arcs in ratio 2 : 1 mean the chord subtends 120° at the centre, so its distance from the centre is 10 cos 60° = 5.
3. The line y − 1 = m(x − 7) is at distance 5 from the origin: (1 − 7m)² = 25(1 + m²), i.e. 12m² − 7m − 12 = 0, so m = 4/3 or −3/4.
4. These lines are 4x − 3y − 25 = 0 and 3x + 4y − 25 = 0: statements (3) and (4).

Teacher's Note:
1. A chord subtending 120° at the centre is at distance r cos 60° from it.

Answer: (d) (1) and (3) only

Explanation:
1. Let P = (p, 0). A tangent y = m(x − p) satisfies m²p² = 50(1 + m²), so m² = 50/(p² − 50).
2. Intercepts on the y-axis are ∓mp, so P₁P₂ = 2p√50/√(p² − 50).
3. Area = ½ × P₁P₂ × p = √50 p²/√(p² − 50). Put u = p²: minimise u²/(u − 50). The derivative gives u = 100.
4. p = ±10: statements (1) and (3).

Teacher's Note:
1. Express the base on the y-axis in terms of p and minimise with u = p².

Answer: (d) All of (1), (2), (3) and (4)

Explanation:
1. The point lies on the circle's diameter, so a² ≤ 4. Let f(x) = x² − 4x − a².
2. f(−1) = 5 − a² > 0 and f(0) = −a² ≤ 0, so one root is in (−1, 0].
3. f(4) = −a² ≤ 0 and f(5) = 5 − a² > 0, so the other root is in [4, 5), the only root in [2, 5].
4. So the roots are distinct, both greater than −1 and both less than 5: all four statements are correct.

Teacher's Note:
1. Check the sign of f at −1, 0, 4 and 5, using a² ≤ 4.

Answer: (b) (1), (2) and (4) only

Explanation:
1. The circle has centre (r, h) and radius √(r² + h² − h²) = r.
2. Perpendicular tangents from the origin mean the origin lies on the director circle: distance of centre from the origin = r√2.
3. r² + h² = 2r², so h² = r², i.e. h = ± r.
4. Statements (1), (2) and (4) are correct.

Teacher's Note:
1. Perpendicular tangents come from a point on the director circle.

Answer: (b) (1) and (2) only

Explanation:
1. The circle passes through the origin, so it is x² + y² + 2gx + 2fy = 0 with intercepts 2|g| and 2|f| on the axes.
2. So |g| = a and |f| = b. The tangent at the origin is gx + fy = 0, i.e. ±ax ± by = 0.
3. This gives ax + by = 0 or ax − by = 0 (for different circles): statements (1) and (2).

Teacher's Note:
1. For a circle through the origin, the tangent there is gx + fy = 0.

Answer: (d) (1), (3) and (4) only

Explanation:
1. OA² = 1, OB² = 2. BC is a chord of C₂ through A. If d is the distance of the origin from the chord, BC² = 4(2 − d²), and 0 ≤ d ≤ 1, so BC² ∈ [4, 8].
2. OA² + OB² + BC² = 3 + BC² ∈ [7, 11]: statement (1) is true, (2) is false.
3. If M is the mid-point of AB, B = 2M − A lies on x² + y² = 2: |2M − A|² = 2, so M lies on the circle with centre A/2 and radius 1/√2.
4. Area = π/2: statements (3) and (4) are true.

Teacher's Note:
1. BC is a chord of C₂ through A. Use the distance of the centre from the chord.
2. B = 2M − A.

Answer: (a) (1), (2) and (3) only

Explanation:
1. The centre lies on the perpendicular bisector x = 2 of the segment joining (−2, 5) and (6, 5), and on x − 3y + 1 = 0, so it is (2, 1).
2. The two given vertices cannot be ends of a diagonal (the mid-point (2, 5) is not on the given diameter), so they are adjacent. The other vertices are their reflections in the centre: (−2, −3) and (6, −3).
3. The sides are 8 and 8, so the area is 64 (a square).
4. The sides lie on x = −2, x = 6, y = 5 and y = −3, not on x = 5 and y = 6: statement (4) is false.

Teacher's Note:
1. Find the centre first, then reflect the vertices in it.

Answer: (c) (3) and (4) only

Explanation:
1. The radii are 1, 1 − d, 1 − 2d with d > 0 (1 is the biggest).
2. The distance of the line from the centre is 1/√2 ≈ 0.707. The line cuts every circle in two points only if the smallest radius exceeds this: 1 − 2d > 0.707, so d < 0.146.
3. d = 0.01 and d = 0.1 satisfy this; 0.4 and 0.6 do not (for 0.4 the smallest radius is 0.2): statements (3) and (4).

Teacher's Note:
1. The smallest circle is the hardest to cut: its radius must exceed the distance of the line from the centre.

Answer: (a) (2) and (3) only

Explanation:
1. Let the circle have centre (h, k) and radius R. The line 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 4ℓ² − 5m² + 6ℓ + 1 = 0: hk = 0, 2h = 6, 2k = 0 so h = 3, k = 0. Then h² − R² = 4 gives R² = 5 (and k² − R² = −5 agrees).
4. Centre (3, 0), radius √5: statements (2) and (3).

Teacher's Note:
1. Write the tangency condition in terms of ℓ and m and compare coefficients.

Answer: (a) (2) and (4) only

Explanation:
1. The common chord is the longest chord of C₁, i.e. a diameter, so it passes through the origin along y = 3x/4 with half-length 4.
2. The centre of C₂ is on the perpendicular to the chord through the origin at distance √(25 − 16) = 3.
3. The unit vector perpendicular to the slope 3/4 is ±(3/5, −4/5), so the centre is ±(9/5, −12/5): (9/5, −12/5) and (−9/5, 12/5).
4. Statements (2) and (4).

Teacher's Note:
1. The centre of C₂ is on the perpendicular bisector of the chord, which passes through the origin.

Answer: (b) (1) and (3) only

Explanation:
1. Circle 1: centre (5, −8), radius |r| (since 25 + 64 − 89 + r² = r²). Circle 2: centre (−3, 7), radius √(9 + 49 − 42) = 4. The distance between the centres is √(64 + 225) = 17.
2. The circles intersect at two points when |r − 4| < 17 < r + 4, i.e. 13 < |r| < 21 (r enters the equation only through r², so r may be negative as well). The integers are ±14, ±15, ..., ±20: 14 values, so statement (1) is true and (2) is false.
3. r = 13: 17 = 13 + 4, external contact: 3 common tangents. Statement (3) is true.
4. r = 21: 17 = 21 − 4, internal contact: only 1 common tangent. Statement (4) is false.

Teacher's Note:
1. Compare the distance of the centres (17) with |r| + 4 and ||r| − 4|.

Answer: (c) (1), (3) and (4) only

Explanation:
1. S₁: centre (0, 0), radius 2. S₂: centre (1, 2), radius 1. The distance between the centres is √5 and 2² + 1² = 5, so the circles are orthogonal: statement (1) is true.
2. Radical axis: x + 2y = 4. Putting x = 4 − 2y in S₁: 5y² − 16y + 12 = 0, so y = 2 or 6/5. The points are (0, 2) and (8/5, 6/5), so statement (2) is false.
3. Chord = √((8/5)² + (4/5)²) = 4/√5 (also 2r₁r₂/d = 4/√5): statement (3) is true.
4. S₁(2, 3) = 9 > 0 and S₂(2, 3) = 1 > 0: statement (4) is true.

Teacher's Note:
1. For orthogonal circles, d² = r₁² + r₂².
2. Find the intersection points from the radical axis.

Answer: (d) (1) and (2) only

Explanation:
1. The centres lie on the normal to the tangent through (1, 2), at distance 5 on either side: (1, 2) ± 5(4, 3)/5 = (5, 5) or (−3, −1).
2. Centre (5, 5): (x − 5)² + (y − 5)² = 25, i.e. x² + y² − 10x − 10y + 25 = 0.
3. Centre (−3, −1): (x + 3)² + (y + 1)² = 25, i.e. x² + y² + 6x + 2y − 15 = 0.
4. Statements (1) and (2).

Teacher's Note:
1. The unit normal to 4x + 3y = 10 is (4, 3)/5.

Answer: (b) (2) and (4) only

Explanation:
1. A circle of radius a touching both circles externally has its centre at distance 2a from (0, 0) and from (2a, 0), so the centre is (a, ±√3 a).
2. Equation: (x − a)² + (y ∓ √3 a)² = a², i.e. x² + y² − 2ax ∓ 2√3 ay + 3a² = 0.
3. Statements (2) and (4) have these forms; (1) and (3) have centres with x = −a.

Teacher's Note:
1. The centre is at distance a + a = 2a from each given centre.

Answer: (a) (1) and (4) only

Explanation:
1. The k-dependent term is −3ky, so the fixed points satisfy y = 0 and x² − 2x − 2 = 0.
2. x = 1 ± √3, giving (1 + √3, 0) and (1 − √3, 0): statements (1) and (4).

Teacher's Note:
1. Collect the terms containing the parameter k.

Answer: (a) (2), (3) and (4) only

Explanation:
1. Adding the two equations: (a + a′)(x² + y²) − 2(g + g′)x − 2(f + f′)y + 2c = 0, which is a circle through A, B, C, D (no xy term, equal coefficients of x² and y²).
2. Its centre is ((g + g′)/(a + a′), (f + f′)/(a + a′)) = P. So P is the centre and PA = PB = PC = PD.
3. P is the centre, not a point on the circle, so statement (1) is false.
4. Equal lengths form both a G.P. and an A.P. (common ratio 1, common difference 0), and PA² + PB² + PC² = 3PD²: statements (2), (3) and (4) are true.

Teacher's Note:
1. Add the two given equations and identify the resulting conic.

Answer: (a) 3

Explanation:
1. A chord subtending the angle α at the centre is at distance 2cos(α/2) from the centre.
2. The chords are on opposite sides of the centre: 2cos(π/2k) + 2cos(π/k) = √3 + 1.
3. k = 3 satisfies this: 2cos 30° + 2cos 60° = √3 + 1. The left side increases with k, so k = 3 is the only solution (a same-side arrangement is impossible, as the difference never reaches √3 + 1).
4. [k] = 3.

Teacher's Note:
1. Distance of a chord from the centre = r cos(half the angle).
2. Test k = 3.

Answer: (b) (−4, 0)

Explanation:
1. Touching the y-axis at (0, 2): the centre is (h, 2) and the radius |h|.
2. Through (−1, 0): (−1 − h)² + 4 = h², so 1 + 2h + 4 = 0 and h = −5/2.
3. Circle: (x + 5/2)² + (y − 2)² = 25/4. At (−4, 0): 9/4 + 4 = 25/4, so it lies on the circle.
4. The other points do not: (−3/2, 0) gives 1 + 4 = 5, (−5/2, 2) is the centre (value 0), and (−3/2, 5/2) gives 1 + 1/4.

Teacher's Note:
1. The centre has the same y-coordinate as the point of contact.

Answer: (c) 2

Explanation:
1. The centre (0, 0) gives 2x − 3y − 1 = −1 < 0, so the smaller part is where 2x − 3y − 1 > 0 (and x² + y² < 6).
2. (2, 3/4): 4 − 9/4 − 1 = 3/4 > 0 and 4 + 9/16 < 6: inside the smaller part.
3. (5/2, 3/4): 25/4 + 9/16 > 6, outside the circle. (1/4, −1/4): 1/2 + 3/4 − 1 = 1/4 > 0 and 1/8 < 6: inside. (1/8, 1/4): 1/4 − 3/4 − 1 < 0, in the larger part.
4. Two points.

Teacher's Note:
1. The smaller part is the side of the line not containing the centre.

Answer: (c) 20(x² + y²) − 36x + 45y = 0

Explanation:
1. Let the mid-point be (h, k). The chord of contact is hx + ky = h² + k², which is also xx₁ + yy₁ = 9 for the point (x₁, y₁).
2. Comparing: x₁ = 9h/(h² + k²), y₁ = 9k/(h² + k²).
3. This point lies on 4x − 5y = 20: 36h − 45k = 20(h² + k²).
4. Locus: 20(x² + y²) − 36x + 45y = 0.

Teacher's Note:
1. Write the chord with a given mid-point as T = S₁ and compare it with the chord of contact from (x₁, y₁).

Answer: (c) (2) and (4) only

Explanation:
1. The common tangent at M meets T at the mid-point of PQ (equal tangent lengths), so M lies on the circle with diameter PQ: centre (0, 1), radius² = 40, i.e. x² + (y − 1)² = 40, with P and Q themselves excluded.
2. (−2, 7) is P, which is excluded, so statement (1) is false. (0, 3/2) gives 1/4 ≠ 40, so it is not on E₁ and statement (4) is true.
3. The mid-points of chords of E₁ through R(1, 1) lie on the circle with diameter joining the centre (0, 1) and R: (x − 1/2)² + (y − 1)² = 1/4, except for chords with P or Q as an end.
4. (1/2, 1) is the centre of that circle, not on it: statement (3) is false. (4/5, 7/5) is on it, but it is the mid-point of the chord through R and P (slope −2), which has the excluded end P, so it does not lie in E₂: statement (2) is true.

Teacher's Note:
1. The locus of M is a circle on PQ as diameter, minus P and Q.
2. Mid-points of chords through a fixed point lie on a circle on (centre, point) as diameter.