CBSE Class 11 Mathematics Conic Sections Competitive Exam MCQs Set 01

Answer: (b) (1, −2)

Explanation:
1. Compare with x² + y² + 2gx + 2fy + c = 0: 2g = −2 and 2f = 4, so g = −1, f = 2.
2. Centre = (−g, −f) = (1, −2).

Teacher's Note:
1. The centre is (−g, −f), i.e. minus half the coefficients of x and y.
2. Make sure the coefficients of x² and y² are both 1 before reading off g and f.

Answer: (b) (−3, −5)

Explanation:
1. Centre = \( \left(\frac{3}{2}, -4\right) \), which is the mid-point of the diameter.
2. If the other end is (x, y), then \( \frac{6 + x}{2} = \frac{3}{2} \) and \( \frac{-3 + y}{2} = -4 \).
3. So x = −3 and y = −5; the other extremity is (−3, −5).

Teacher's Note:
1. The centre is the mid-point of every diameter, so the other end = 2 × centre − given end.
2. Check that (6, −3) actually lies on the circle: 36 + 9 − 18 − 24 − 3 = 0.

Answer: (d) \( \frac{1}{2} \)

Explanation:
1. Divide by 2: x² + y² + \( \frac{3}{2} \)x + 2y − \( \frac{1}{2} \) = 0, so the centre is \( \left(-\frac{3}{4}, -1\right) \).
2. A diameter passes through the centre: \( -1 = 2\left(-\frac{3}{4}\right) + K \).
3. K = −1 + \( \frac{3}{2} \) = \( \frac{1}{2} \).

Teacher's Note:
1. Every diameter passes through the centre, so substitute the centre in the line.
2. Always divide by the common coefficient of x² and y² before finding the centre.

Answer: (b) a circle

Explanation:
1. By the section formula, P = \( \left(\frac{2 \cdot 10\cos\theta + 3 \cdot 5}{5}, \frac{2 \cdot 10\sin\theta + 0}{5}\right) \) = (3 + 4 cos θ, 4 sin θ).
2. So x − 3 = 4 cos θ and y = 4 sin θ.
3. Squaring and adding: (x − 3)² + y² = 16, which is a circle with centre (3, 0) and radius 4.

Teacher's Note:
1. To find a locus, write x and y in terms of θ and eliminate θ using cos²θ + sin²θ = 1.
2. An equation of the form (x − h)² + (y − k)² = r² is always a circle.

Answer: (d) 4

Explanation:
1. A circle of radius r touching both axes has its centre at a distance r from each axis.
2. So the centre is (±r, ±r), one in each quadrant.
3. Hence there are 4 such circles.

Teacher's Note:
1. A circle touching the x-axis has radius = |y-coordinate of centre|; touching the y-axis, radius = |x-coordinate|.
2. The four circles are (x ± r)² + (y ± r)² = r².

Answer: (b) \( \frac{3}{4} \)

Explanation:
1. For a circle, the coefficients of x² and y² must be equal.
2. Coefficient of x² = \( \frac{\lambda}{3} \), coefficient of y² = \( \frac{1}{4} \).
3. \( \frac{\lambda}{3} = \frac{1}{4} \) gives λ = \( \frac{3}{4} \).

Teacher's Note:
1. A second-degree equation is a circle only when x² and y² have equal coefficients and there is no xy term.
2. With λ = 3/4 the equation becomes (x + 1)² + (y + 2)² = 4, a circle of radius 2.

Answer: (c) \( \frac{25}{2} \)

Explanation:
1. Divide by 2: x² + y² − 3x + 4y + \( \frac{k}{2} \) = 0, so g = \( -\frac{3}{2} \), f = 2, c = \( \frac{k}{2} \).
2. Point circle: g² + f² − c = 0, i.e. \( \frac{9}{4} + 4 - \frac{k}{2} = 0 \).
3. \( \frac{k}{2} = \frac{25}{4} \), so k = \( \frac{25}{2} \).

Teacher's Note:
1. A point circle has radius zero, so g² + f² − c = 0.
2. Forgetting to divide by 2 first is the most common mistake here.

Answer: (d) a = b ≠ 0, h = 0, g² + f² − ac > 0

Explanation:
1. For a circle we need equal coefficients of x² and y² (a = b ≠ 0) and no xy term (h = 0).
2. Dividing by a: x² + y² + \( \frac{2g}{a} \)x + \( \frac{2f}{a} \)y + \( \frac{c}{a} \) = 0.
3. Real radius needs \( \frac{g^2}{a^2} + \frac{f^2}{a^2} - \frac{c}{a} > 0 \), i.e. g² + f² − ac > 0.

Teacher's Note:
1. The radius condition must be applied only after making the coefficient of x² equal to 1.
2. Option (c) is correct only in the special case a = 1.

Answer: (d) a point if λ = 0

Explanation:
1. Complete the squares: 2(x² − 4x) + 3(y² − 6y) + 35 = λ.
2. 2(x − 2)² − 8 + 3(y − 3)² − 27 + 35 = λ, i.e. 2(x − 2)² + 3(y − 3)² = λ.
3. If λ = 0, the only solution is x = 2, y = 3, a single point.

Teacher's Note:
1. Since the coefficients of x² and y² are different (2 and 3), the curve can never be a circle.
2. A sum of squares equal to zero forces each square to be zero, giving a single point.

Answer: (b) x² + y² − 4x + 2y − 4 = 0

Explanation:
1. (x − 2)² + (y + 1)² = 9
2. x² − 4x + 4 + y² + 2y + 1 = 9
3. x² + y² − 4x + 2y − 4 = 0

Teacher's Note:
1. Use (x − h)² + (y − k)² = r² and expand carefully.
2. Check: centre (−g, −f) = (2, −1) and radius √(4 + 1 + 4) = 3.

Answer: (c) x² + y² + 4x + 6y + 9 = 0

Explanation:
1. Touching the y-axis means radius = distance of the centre from the y-axis = |−2| = 2.
2. (x + 2)² + (y + 3)² = 4
3. x² + y² + 4x + 6y + 9 = 0

Teacher's Note:
1. Touching the y-axis: r = |x-coordinate of centre|; touching the x-axis: r = |y-coordinate|.
2. Check: putting x = 0 gives (y + 3)² = 0, a single point of contact with the y-axis.

Answer: (c) x² + y² − 6x − 10y + 9 = 0

Explanation:
1. Touching the x-axis at (3, 0): centre (3, k), radius |k|.
2. Equation: x² + y² − 6x − 2ky + 9 = 0; intercept on y-axis = \( 2\sqrt{f^2 - c} = 2\sqrt{k^2 - 9} \) = 8.
3. k² − 9 = 16, so k = 5 (the intercept is on the positive y-axis).
4. Equation: x² + y² − 6x − 10y + 9 = 0.

Teacher's Note:
1. For a circle touching the x-axis at (a, 0), the centre is (a, k) and the radius is |k|.
2. Check: putting x = 0 gives y² − 10y + 9 = 0, so y = 1, 9 and the intercept is 8.

Answer: (b) x² + y² − 4x − 6y − 87 = 0

Explanation:
1. Centre of the first circle = (−4, −5).
2. Second circle: x² + y² − 4x − 6y − \( \frac{9}{2} \) = 0, centre (2, 3).
3. Radius² = (2 + 4)² + (3 + 5)² = 36 + 64 = 100.
4. (x − 2)² + (y − 3)² = 100 gives x² + y² − 4x − 6y − 87 = 0.

Teacher's Note:
1. Concentric circles have the same centre; only the constant term changes.
2. The radius is the distance from the common centre to the given point.

Answer: (c) 4x² + 4y² − 12x + 16y − 21 = 0

Explanation:
1. Given circle: x² + y² − 3x + 4y + \( \frac{1}{2} \) = 0, centre \( \left(\frac{3}{2}, -2\right) \), r² = \( \frac{9}{4} + 4 - \frac{1}{2} = \frac{23}{4} \).
2. Double area means double r², so the new r² = \( \frac{23}{2} \).
3. \( \left(x - \frac{3}{2}\right)^2 + (y + 2)^2 = \frac{23}{2} \) gives x² + y² − 3x + 4y − \( \frac{21}{4} \) = 0.
4. Multiplying by 4: 4x² + 4y² − 12x + 16y − 21 = 0.

Teacher's Note:
1. Area is πr², so doubling the area doubles r², not r.
2. Keep the same centre and change only the constant term.

Answer: (a) x² + y² − 12y + 11 = 0

Explanation:
1. Let the centre be (0, k), k > 0. Then 3² + (2 − k)² = 25.
2. (2 − k)² = 16, so k = 6 or k = −2; take k = 6.
3. x² + (y − 6)² = 25 gives x² + y² − 12y + 11 = 0.

Teacher's Note:
1. A centre on the y-axis has the form (0, k).
2. The condition 'first quadrant' (positive y-axis) rejects k = −2.

Answer: (b) 3(x² + y²) − 25x = 0

Explanation:
1. Touching the y-axis at the origin: centre (h, 0), radius |h|, so the circle is x² + y² − 2hx = 0.
2. Passing through (3, 4): 9 + 16 − 6h = 0, so h = \( \frac{25}{6} \).
3. x² + y² − \( \frac{25}{3} \)x = 0, i.e. 3(x² + y²) − 25x = 0.

Teacher's Note:
1. A circle touching the y-axis at the origin has its centre on the x-axis.
2. Such a circle always has the form x² + y² − 2hx = 0.

Answer: (a) −4

Explanation:
1. Concentric with the given circle: g = −1, f = 2.
2. (4, −2) lies on it: 16 + 4 − 8 − 8 + c = 0.
3. c = −4.

Teacher's Note:
1. Concentric circles share g and f; only c differs.
2. Substitute the point after fixing g and f to find c.

Answer: (b) 15(x² + y²) − 94x + 18y + 55 = 0

Explanation:
1. Let the circle be x² + y² + 2gx + 2fy + c = 0: 5 + 2g − 4f + c = 0 and 25 + 8g − 6f + c = 0.
2. Subtracting: 20 + 6g − 2f = 0; centre on the line: −3g − 4f = 7.
3. Solving: g = \( -\frac{47}{15} \), f = \( \frac{3}{5} \), and then c = \( \frac{11}{3} \).
4. Multiplying by 15: 15(x² + y²) − 94x + 18y + 55 = 0.

Teacher's Note:
1. Two points give two equations and the line through the centre gives the third.
2. Subtracting the two point equations removes c and simplifies the work.

Answer: (c) \( \sqrt{3}(x^2 + y^2) - 2y - \sqrt{3} = 0 \)

Explanation:
1. Third vertex = (0, √3). For an equilateral triangle the circumcentre is the centroid \( \left(0, \frac{1}{\sqrt{3}}\right) \).
2. Radius = distance to (1, 0) = \( \sqrt{1 + \frac{1}{3}} = \frac{2}{\sqrt{3}} \).
3. \( x^2 + \left(y - \frac{1}{\sqrt{3}}\right)^2 = \frac{4}{3} \) gives \( x^2 + y^2 - \frac{2}{\sqrt{3}}y - 1 = 0 \).
4. Multiplying by √3: \( \sqrt{3}(x^2 + y^2) - 2y - \sqrt{3} = 0 \).

Teacher's Note:
1. In an equilateral triangle the centroid, circumcentre and orthocentre coincide.
2. The circumradius of an equilateral triangle of side a is a/√3; here a = 2.

Answer: (d) 2x² + 2y² − 5x + 11y − 3 = 0

Explanation:
1. Let x² + y² + 2gx + 2fy + c = 0: 9 + 6g + c = 0, 37 + 2g − 12f + c = 0, 17 + 8g − 2f + c = 0.
2. Solving: g = \( -\frac{5}{4} \), f = \( \frac{11}{4} \), c = \( -\frac{3}{2} \).
3. x² + y² − \( \frac{5}{2} \)x + \( \frac{11}{2} \)y − \( \frac{3}{2} \) = 0, i.e. 2x² + 2y² − 5x + 11y − 3 = 0.

Teacher's Note:
1. With options given, substituting the three points in each option is a quick check.
2. For (3, 0) in option (d): 18 − 15 − 3 = 0; the other two points also satisfy it.

Answer: (d) \( 2\sqrt{2} \)

Explanation:
1. The segment from (1, 2) to (5, 2) is horizontal and from (5, 2) to (5, −2) is vertical, so the angle at (5, 2) is 90°.
2. Hence (1, 2) and (5, −2) are the ends of a diameter.
3. Diameter = \( \sqrt{16 + 16} = 4\sqrt{2} \), so the radius = \( 2\sqrt{2} \).

Teacher's Note:
1. The angle in a semicircle is a right angle, so the hypotenuse of a right triangle is a diameter of its circumcircle.
2. Spotting the right angle saves solving three simultaneous equations.

Answer: (a) x² + y² + 2ax + 2cy + b + d = 0

Explanation:
1. Let A = (x₁, y₁), B = (x₂, y₂). Circle on AB as diameter: (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0.
2. (x − x₁)(x − x₂) = x² + 2ax + b, since x₁, x₂ are the roots of x² + 2ax + b = 0.
3. Similarly (y − y₁)(y − y₂) = y² + 2cy + d.
4. Adding: x² + y² + 2ax + 2cy + b + d = 0.

Teacher's Note:
1. (x − x₁)(x − x₂) is exactly the quadratic whose roots are x₁ and x₂.
2. This trick avoids finding the roots at all.

Answer: (c) x² + y² − 5x − 5y = 0

Explanation:
1. The circle passes through (0, 0), (5, 0) and (0, 5).
2. From the origin, c = 0; from (5, 0): 25 + 10g = 0, g = \( -\frac{5}{2} \); from (0, 5): f = \( -\frac{5}{2} \).
3. Equation: x² + y² − 5x − 5y = 0.

Teacher's Note:
1. A circle through the origin has no constant term.
2. Since the angle at the origin is 90°, the points (5, 0) and (0, 5) are ends of a diameter.

Answer: (c) \( 4\sqrt{13} \)

Explanation:
1. Circle: (x + 4)(x − 12) + (y − 3)(y + 1) = 0, i.e. x² + y² − 8x − 2y − 51 = 0.
2. Here f = −1, c = −51.
3. Intercept on the y-axis = \( 2\sqrt{f^2 - c} = 2\sqrt{1 + 51} = 4\sqrt{13} \).

Teacher's Note:
1. The y-intercept uses f (the y-coefficient) and the x-intercept uses g.
2. Alternatively, put x = 0 and find the difference of the roots of y² − 2y − 51 = 0.

Answer: (d) 1

Explanation:
1. Put x = 0: y² − 7y + 12 = 0, so y = 3 or y = 4.
2. Intercept = 4 − 3 = 1.

Teacher's Note:
1. The intercept on an axis is the distance between the two points where the circle cuts it.
2. Formula check: \( 2\sqrt{f^2 - c} = 2\sqrt{\frac{49}{4} - 12} = 1 \).

Answer: (d) (1, 0), (2, 0)

Explanation:
1. Put y = 0: x² − 3x + 2 = 0.
2. (x − 1)(x − 2) = 0, so x = 1, 2.
3. Points: (1, 0) and (2, 0).

Teacher's Note:
1. Points on the x-axis have y = 0; substitute and solve the quadratic in x.
2. Two distinct real roots mean the circle cuts the x-axis at two points.

Answer: (d) (2π/3, 4π/3)

Explanation:
1. Interior point: (1 + cos θ)² + sin²θ − 1 < 0.
2. 1 + 2 cos θ + cos²θ + sin²θ − 1 < 0, i.e. 1 + 2 cos θ < 0, so cos θ < \( -\frac{1}{2} \).
3. In [0, 2π] this holds for 2π/3 < θ < 4π/3.

Teacher's Note:
1. A point (x₁, y₁) is inside the circle S = 0 when S₁ < 0.
2. cos θ = −1/2 at θ = 2π/3 and 4π/3; cos θ is less than −1/2 between them.

Answer: (c) x² + y² − 2x − 4y = 0

Explanation:
1. Points of intersection: x² + 4x² = 10x, so 5x(x − 2) = 0, x = 0 or 2.
2. The ends of the chord are (0, 0) and (2, 4).
3. Circle on this diameter: x(x − 2) + y(y − 4) = 0, i.e. x² + y² − 2x − 4y = 0.

Teacher's Note:
1. Find the ends of the chord by solving the line and the circle together.
2. Then use the diameter form (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0.

Answer: (a) x² + y² + 3x − 3y − a² + 9 = 0

Explanation:
1. Substituting y = x + 3: 2x² + 6x + 9 − a² = 0, so x₁ + x₂ = −3 and x₁x₂ = \( \frac{9 - a^2}{2} \).
2. y₁ + y₂ = (x₁ + x₂) + 6 = 3 and y₁y₂ = x₁x₂ + 3(x₁ + x₂) + 9 = \( \frac{9 - a^2}{2} \).
3. Circle: x² − (x₁ + x₂)x + x₁x₂ + y² − (y₁ + y₂)y + y₁y₂ = 0.
4. x² + y² + 3x − 3y + 9 − a² = 0.

Teacher's Note:
1. You do not need the actual points; sums and products of roots are enough for the diameter form.
2. Check: the centre (−3/2, 3/2) lies on y = x + 3, as the mid-point of AB must.

Answer: (a) 9(x² + y²) = 4a²

Explanation:
1. Let A = (p, 0) and B = (0, q). Since ∠AOB = 90°, AB is a diameter, so p² + q² = (2a)² = 4a².
2. Centroid of OAB: x = \( \frac{p}{3} \), y = \( \frac{q}{3} \), so p = 3x and q = 3y.
3. 9x² + 9y² = 4a², i.e. 9(x² + y²) = 4a².

Teacher's Note:
1. An angle of 90° at the origin means AB subtends a right angle on the circle, so AB is a diameter.
2. For a locus, express the given relation in terms of the moving point's coordinates.

Answer: (b) the area of the circle is larger

Explanation:
1. Let the perimeter be L. Circle: r = \( \frac{L}{2\pi} \), area = \( \frac{L^2}{4\pi} \).
2. Square: side \( \frac{L}{4} \), area = \( \frac{L^2}{16} \).
3. Since 4π < 16, \( \frac{L^2}{4\pi} > \frac{L^2}{16} \), so the circle has the larger area.

Teacher's Note:
1. For a fixed perimeter, the circle encloses the largest possible area.
2. Compare 1/(4π) ≈ 0.080 with 1/16 = 0.0625.

Answer: (b) x² + 8x + (y − 3)² = 0

Explanation:
1. Let M = (h, k). Since AM = 2AB, B is the mid-point of AM: B = \( \left(\frac{h}{2}, \frac{k + 3}{2}\right) \).
2. B lies on the circle: \( \frac{h^2}{4} + 2h + \left(\frac{k - 3}{2}\right)^2 = 0 \).
3. Multiplying by 4: h² + 8h + (k − 3)² = 0, so the locus is x² + 8x + (y − 3)² = 0.

Teacher's Note:
1. Express the known point B in terms of the moving point M and substitute in the circle.
2. AM = 2AB simply says B is the mid-point of AM.

Answer: (b) x² − ax + b = 0

Explanation:
1. Circle: (x − 0)(x − a) + (y − 1)(y − b) = 0.
2. On the x-axis, y = 0: x² − ax + b = 0.

Teacher's Note:
1. Put y = 0 in the circle to get the x-coordinates of the points where it meets the x-axis.
2. (0 − 1)(0 − b) = b, so the constant term is +b.