JEE Mathematics Circle MCQs Set 06

Answer: (d) x² + y² − 2x − 2y − 11 = 0

Explanation:
1. A normal passes through the centre, so the segment joining (3, 4) and (−1, −2) is a diameter.
2. Circle: (x − 3)(x + 1) + (y − 4)(y + 2) = 0.
3. Expanding: x² − 2x − 3 + y² − 2y − 8 = 0, i.e. x² + y² − 2x − 2y − 11 = 0.

Teacher's Note:
1. A normal meets the circle at the two ends of a diameter.
2. Once the end-points of a diameter are known, use the diameter form of the circle.

Answer: (c) (1, 4)

Explanation:
1. Centre (3, −1). The mid-point of the chord is the foot of the perpendicular from the centre to the line.
2. Value of the line at the centre: 2(3) − 5(−1) + 18 = 29. Foot = (3, −1) − (29/29)(2, −5) = (1, 4).
3. Check: 2(1) − 5(4) + 18 = 0, so (1, 4) lies on the line.

Teacher's Note:
1. The line from the centre to the chord's mid-point is perpendicular to the chord.
2. Use the foot-of-perpendicular formula and then verify the point lies on the line.

Answer: (b) x² + y² = 2

Explanation:
1. The origin is the centre. A chord subtending 90° at the centre has half-angle 45°, so its distance from the centre is \( 2\cos45^\circ=\sqrt2 \).
2. The mid-point of the chord is at this distance from the origin.
3. Locus: x² + y² = 2.

Teacher's Note:
1. The mid-point of a chord is the foot of the perpendicular from the centre.
2. For central angle 2φ, the distance of the chord from the centre is r cos φ.

Answer: (d) \( \left(\tfrac12,\tfrac14\right) \)

Explanation:
1. For a point (h, k) on the line, k = 4 − 2h. Its chord of contact is hx + ky = 1.
2. hx + (4 − 2h)y = 1, i.e. h(x − 2y) + (4y − 1) = 0 for every h.
3. So x − 2y = 0 and 4y − 1 = 0, giving y = 1/4, x = 1/2.

Teacher's Note:
1. Write the chord as h times one expression plus another, so the fixed point satisfies both expressions.
2. The fixed point is the pole of the line 2x + y = 4 with respect to the circle.

Answer: (d) 2x − 5y + 11 = 0

Explanation:
1. The line from the centre to the mid-point of a chord is perpendicular to the chord.
2. The chord has slope −5/2, so the line through (2, 3) perpendicular to it has slope 2/5.
3. y − 3 = (2/5)(x − 2), i.e. 2x − 5y + 11 = 0.

Teacher's Note:
1. All the centres lie on the perpendicular to the chord through its mid-point.
2. Check that (2, 3) satisfies the answer: 4 − 15 + 11 = 0.

Answer: (c) x² + y² − 2x − 2y = 0

Explanation:
1. Let the mid-point be (h, k). The chord is hx + ky = h² + k² (= s).
2. Homogenise x² − 2(x + y) = 0 using 1 = (hx + ky)/s: \( x^2-2(x+y)\frac{hx+ky}{s}=0 \).
3. The lines are perpendicular if the sum of the coefficients of x² and y² is zero: (1 − 2h/s) + (−2k/s) = 0, so s = 2h + 2k.
4. h² + k² − 2h − 2k = 0. Locus: x² + y² − 2x − 2y = 0.

Teacher's Note:
1. To find the angle subtended at the origin, make the curve homogeneous with the chord equation.
2. A pair of lines ax² + 2hxy + by² = 0 is perpendicular when a + b = 0.

Answer: (b) 1

Explanation:
1. Centres (−2, 2) and (2, −1); radii 7 and 2.
2. Distance between centres: \( \sqrt{16+9}=5 \). Also r₁ − r₂ = 5.
3. Since the distance equals r₁ − r₂, the circles touch internally, so there is exactly one common tangent.

Teacher's Note:
1. Compare the distance between centres with r₁ + r₂ and |r₁ − r₂|.
2. Distance = |r₁ − r₂| means internal touching and one common tangent.

Answer: (b) 5x + 12y + 19 = 0

Explanation:
1. Circle 1: centre (2, 3), radius 5. Circle 2: centre (−3, −9), radius \( \sqrt{9+81-26}=8 \).
2. Distance between centres = √(25 + 144) = 13 = 5 + 8, so they touch externally.
3. The common tangent at the contact point is the radical axis S₁ − S₂ = 0: −10x − 24y − 38 = 0, i.e. 5x + 12y + 19 = 0.

Teacher's Note:
1. First decide how the circles touch, then use S₁ − S₂ = 0 for the tangent at the point of contact.
2. Check: the line is perpendicular to the line of centres, whose direction is (5, 12).

Answer: (c) x² + y² − 16x − 18y − 4 = 0

Explanation:
1. Let the required circle be x² + y² + 2Gx + 2Fy + C = 0. Orthogonality with x² + y² + 2gx + 2fy + c = 0 requires 2Gg + 2Ff = C + c.
2. With g, f, c = (−1, 3/2, −7), (5/2, −5/2, 9), (7/2, −9/2, 29) we get −2G + 3F = C − 7, 5G − 5F = C + 9, 7G − 9F = C + 29.
3. Solving: G = −8, F = −9, C = −4.
4. Circle: x² + y² − 16x − 18y − 4 = 0.

Teacher's Note:
1. Write the orthogonality condition for each given circle and solve the three linear equations.
2. Check the first condition: 2(−8)(−1) + 2(−9)(3/2) = 16 − 27 = −11 = −4 + (−7).

Answer: (a) 18

Explanation:
1. With d the distance between centres: L_int² = d² − (r₁ + r₂)² = 49 and L_ext² = d² − (r₁ − r₂)² = 121.
2. Subtract: (r₁ + r₂)² − (r₁ − r₂)² = 121 − 49 = 72.
3. 4r₁r₂ = 72, so r₁r₂ = 18.

Teacher's Note:
1. Subtracting the two tangent-length formulas eliminates d.
2. (a + b)² − (a − b)² = 4ab.

Answer: (c) a straight line

Explanation:
1. Let the moving circle be x² + y² + 2Gx + 2Fy + C = 0. It bisects the circumference of x² + y² + 2g₁x + 2f₁y + c₁ = 0 when their common chord (the radical axis) passes through the centre (−g₁, −f₁) of the second circle.
2. Imposing this gives 2g₁G + 2f₁F − C = 2g₁² + 2f₁² − c₁, which is linear in G, F, C.
3. For x² + y² = 4 we have g₁ = f₁ = 0 and c₁ = −4, so the condition reads −C = 0 − (−4) = 4, that is C = −4. For the other circle (g₁ = −1, f₁ = 3, c₁ = 1): −2G + 6F − C = 2 + 18 − 1 = 19.
4. Substituting C = −4: −2G + 6F = 15, which is linear in G and F, so the centre (−G, −F) moves on a straight line: 2x − 6y = 15.

Teacher's Note:
1. Bisecting a circumference means the common chord is a diameter of the bisected circle.
2. Because the condition is linear in G and F, the locus of the centre is a line.

Answer: (c) (x² + y² − a²) − 2p(x cos α + y sin α − p) = 0

Explanation:
1. Circle through the intersection of the circle and the chord: x² + y² − a² + λ(x cos α + y sin α − p) = 0.
2. Its centre is (−λcos α/2, −λ sin α/2). For the chord to be a diameter the centre must lie on the chord: −λ/2 − p = 0, so λ = −2p.
3. Equation: (x² + y² − a²) − 2p(x cos α + y sin α − p) = 0.

Teacher's Note:
1. Use S + λL = 0 for a circle through the common points of a circle and a line.
2. Choose λ so that the centre lies on the line.

Answer: (b) x² + y² + x − 6y + 3 = 0

Explanation:
1. Tangent at (2, 3): 2x + 3y + 2(x + 2) − 3(y + 3) − 3 = 4x − 8 = 0, i.e. x = 2.
2. Circle touching S = 0 at (2, 3): S + λ(4x − 8) = 0.
3. Through (1, 1): S(1, 1) = 1 + 1 + 4 − 6 − 3 = −3 and 4 − 8 = −4, so −3 − 4λ = 0, λ = −3/4.
4. S − (3/4)(4x − 8) = x² + y² + x − 6y + 3 = 0.

Teacher's Note:
1. A circle touching a given circle at a given point belongs to the family S + λT = 0, where T is the tangent there.
2. Substitute the extra point to fix λ.

Answer: (a) 2x² + 2y² − 10x − 5y + 1 = 0

Explanation:
1. Circle touching the line at (1, −1): (x − 1)² + (y + 1)² + k(2x + 3y + 1) = 0, i.e. x² + y² + (2k − 2)x + (3k + 2)y + (2 + k) = 0.
2. Second circle: x(x + 2) + (y − 3)(y + 1) = 0, i.e. x² + y² + 2x − 2y − 3 = 0.
3. Orthogonality: 2g₁g₂ + 2f₁f₂ = c₁ + c₂ gives 2(k − 1)(1) + 2((3k + 2)/2)(−1) = (2 + k) − 3.
4. 2k − 2 − 3k − 2 = k − 1, so k = −3/2.
5. Equation: x² + y² − 5x − (5/2)y + 1/2 = 0, i.e. 2x² + 2y² − 10x − 5y + 1 = 0.

Teacher's Note:
1. Use the family of circles touching a line at a given point, then impose orthogonality.
2. Check that (1, −1) lies on the answer: 2 + 2 − 10 + 5 + 1 = 0.

Answer: (b) \( x^2+y^2+2x-\tfrac32 y=0 \)

Explanation:
1. Tangent at the origin to S = 0: −4x + 3y = 0.
2. Required circle: x² + y² − 8x + 6y + λ(−4x + 3y) = 0.
3. Through (−1, 2): 1 + 4 + 8 + 12 + λ(4 + 6) = 0, so 25 + 10λ = 0 and λ = −5/2.
4. Equation: x² + y² + 2x − (3/2)y = 0.

Teacher's Note:
1. Tangent at the origin is obtained by keeping only the linear terms: −8x/2 + 6y/2 = 0 gives −4x + 3y = 0.
2. Substitute the given point to find the parameter.

Answer: (d) \( \frac{40}{9} \)

Explanation:
1. Take a = 1. A circle touching the y-axis with centre (h, k) has radius |h|, so (1 − h)² + (5 − k)² = h² and (4 − h)² + (1 − k)² = h².
2. Solving gives the two circles: centre (5/2, 3), r = 5/2 and centre (205/18, 29/3), r = 205/18.
3. With d the distance between centres: cos θ = (r₁² + r₂² − d²)/(2r₁r₂) = 9/41.
4. sin θ = 40/41, so tan θ = 40/9.

Teacher's Note:
1. Angle between circles equals the angle between their radii at a common point: use the cosine rule on the triangle of centres and an intersection point.
2. The result does not depend on a, because every length scales with a.

Answer: (d) 1

Explanation:
1. The points lie on the curve xy = 1. Let the circle be (x − p)² + (y − q)² = 16.
2. Substitute y = 1/x and multiply by x²: (x − p)²x² + (1 − qx)² = 16x². This is a quartic in x whose roots are a, b, c, d.
3. Leading coefficient 1; constant term (from (1 − qx)² at x = 0) = 1.
4. Product of the roots abcd = 1/1 = 1.

Teacher's Note:
1. Substitute the hyperbola into the circle and look at the quartic in x.
2. For a degree-4 polynomial the product of roots equals the constant term divided by the leading coefficient.

Answer: (c) x² + 8x + (y − 3)² = 0

Explanation:
1. B is the mid-point of AM. With M = (x, y), B = (x/2, (y + 3)/2).
2. B lies on the circle: x²/4 + 4(x/2) + ((y + 3)/2 − 3)² = 0.
3. x²/4 + 2x + (y − 3)²/4 = 0, so x² + 8x + (y − 3)² = 0.

Teacher's Note:
1. Because AM = 2AB, the point B is the mid-point of AM.
2. Express B in terms of M and substitute in the given circle.

Answer: (b) \( \left(\frac{9}{5},\frac{18}{5}\right) \)

Explanation:
1. Tangent at (1, 2) to C₁: x + 2y = 5. This line is the chord AB of C₂.
2. The chord AB is the chord of contact of C from C(h, k): hx + ky = 9.
3. Compare hx + ky = 9 with x + 2y = 5: h/1 = k/2 = 9/5.
4. C = (9/5, 18/5).

Teacher's Note:
1. Tangents at the ends of a chord meet at the pole of the chord, and the chord is the chord of contact from that point.
2. Match the coefficients to find the pole.

Answer: (d) \( \frac{g^2+f^2-c}{2\sqrt{g^2+f^2}} \)

Explanation:
1. From the origin: gx + fy + c = 0.
2. From (g, f): gx + fy + g(x + g) + f(y + f) + c = 0, i.e. gx + fy + (g² + f² + c)/2 = 0.
3. The lines are parallel; distance = |(g² + f² + c)/2 − c| / √(g² + f²) = (g² + f² − c)/(2√(g² + f²)).

Teacher's Note:
1. Both chords have the same x and y coefficients, so use the distance between parallel lines.
2. Divide the second equation by 2 so the coefficients match the first.

Answer: (b) \( 2\alpha \)

Explanation:
1. Both circles have centre (−g, −f). Let R² = g² + f² − c. The first circle has radius R.
2. Radius of the second: r² = g² + f² − c sin²α − (g² + f²)cos²α = (g² + f² − c) sin²α, so r = R sin α.
3. From P, with PC = R and radius r, sin(half-angle) = r/R = sin α, so the half-angle is α.
4. Angle between the tangents = 2α.

Teacher's Note:
1. Find the radius of the second circle in terms of R, and use sin(half-angle) = radius / distance from centre.
2. P is at distance R from the common centre.

Answer: (b) (x + 2)² + (y − 3)² = 6.25

Explanation:
1. Centre (−2, 3), radius \( \sqrt{4+9+12}=5 \).
2. A chord subtending π/3 at the circumference subtends 2π/3 at the centre, so the distance of the chord from the centre is 5 cos(π/3) = 2.5.
3. The mid-points lie on the circle with the same centre and radius 2.5: (x + 2)² + (y − 3)² = 6.25.

Teacher's Note:
1. Angle at the centre is twice the angle at the circumference.
2. The mid-point of a chord is at distance r cos(half central angle) from the centre.