NDA Circle previous year questions with solutions

Answer: (b) x² + y² = 25

Method. diameters meet at the centre

The diameters x + y = 0 and x − y = 0 intersect at (0, 0), so the centre is the origin.
Diameter 10 gives radius 5.
x² + y² = 25.

Answer: (a) 19

Method. centre (-g, -f) and r² = g² + f² - c

g = 1, f = 3, c = 1.
Centre (a, b) = (-1, -3), so a² + b² = 1 + 9 = 10.
c² = r² = 1 + 9 − 1 = 9.
a² + b² + c² = 10 + 9 = 19.

Answer: (c) 3x − 4y + 6 = 0

Method. a circle through the origin cutting intercepts a, b on the axes has (a, 0), (0, b) as ends of a diameter

The circle passes through O(0, 0), (4, 0) and (0, 6).
The angle at O between the axes is 90°, so the chord joining (4, 0) and (0, 6) is a diameter.
Centre = mid-point = (2, 3).
Check the options: 3(2) − 4(3) + 6 = 0, so the centre lies on 3x − 4y + 6 = 0.
(The other lines give 2, -1 and -13, not 0; taking intercepts the other way round, centre (3, 2), satisfies none of the four lines.)

Answer: (d) 3.5 units

Method. make the x², y² coefficients 1, then r = ⁠√(g²⁠ + f² - c)

Divide by 4: x² + y² - 5x + 3y − 154 = 0.
g = -52, f = 32, c = -154.
r² = 254 + 94 + 154 = 494.
r = 72 = 3.5 units.

Answer: (a) 1

Method. y-intercept = difference of the roots obtained by putting x = 0

Put x = 0: y² - 7y + 12 = 0, so (y − 3)(y − 4) = 0.
The circle cuts the y-axis at (0, 3) and (0, 4).
Intercept = 4 − 3 = 1.

Shortcut. Intercept on y-axis = 2⁠√(f²⁠ - c) = 2⁠√(494⁠ −⁠ 12) = 2 · 12 = 1.

Answer: (c) 2, 17

Method. find the circle through A, B, C and see where it meets the y-axis

Let x² + y² + 2gx + 2fy + c = 0.
A(0, 2): 4 + 4f + c = 0. B(2, 3): 13 + 4g + 6f + c = 0. C(4, 5): 41 + 8g + 10f + c = 0.
B − A: 9 + 4g + 2f = 0; C − B: 28 + 4g + 4f = 0. Subtracting: 2f = -19, f = -192; then g = 52, c = 34.
Circle: x² + y² + 5x − 19y + 34 = 0.
D(0, k) on it: k² - 19k + 34 = 0, (k − 2)(k − 17) = 0, so k = 2 or 17.
(k = 2 makes D coincide with A; k = 17 gives a fourth distinct point.)

Answer: (b) 5⁠√10⁠

Method. with k = 17 from the previous item, the circle through A, B, D is the circle through A, B, C

From the previous item, D(0, 17) (k = 2 would make D the same point as A, and no unique circle would pass through A, B, D).
A, B, C, D are concyclic, so the circle through A, B, D is x² + y² + 5x − 19y + 34 = 0.
r² = 254 + 3614 − 34 = 2504, so r = 5⁠√102.⁠
Diameter = 5⁠√10.⁠

Answer: (a) I and II only

Method. diameter form, then any chord through the centre with ends on the circle is a diameter

(x − 1)(x − 3) + (y − 2)(y − 4) = 0 is the circle on the diameter joining (1, 2) and (3, 4): I is correct.
Centre (2, 3), r² = 1 + 1 = 2.
II: (1, 4) and (3, 2) each lie at distance ⁠√2⁠ from (2, 3), and their mid-point is (2, 3): they are ends of a diameter. Correct.
III: mid-point of (2, 4) and (4, 2) is (3, 3), not the centre; also (4, 2) is at distance ⁠√5⁠ from (2, 3), so it is not even on the circle. Wrong.
I and II only.

Answer: (d) (-58, -24)

Method. centre is equidistant from the three points (intersection of perpendicular bisectors)

Let the centre be (h, k).
Equal distances to (5, -8) and (2, 1): (h − 5)² + (k + 8)² = (h − 2)² + (k − 1)², which gives -6h + 18k + 84 = 0, i.e. h − 3k = 14.
Equal distances to (-2, 9) and (2, 1): (h + 2)² + (k − 9)² = (h − 2)² + (k − 1)², which gives 8h − 16k + 80 = 0, i.e. h − 2k = -10.
Subtracting: k = -24, then h = 14 + 3(-24) = -58.
Centre (-58, -24).

Answer: (d) r > 60

Method. radius = distance from the centre found in the previous item to any of the points

Centre (-58, -24) (previous item); take the point (2, 1).
r² = (2 + 58)² + (1 + 24)² = 3600 + 625 = 4225.
r = 65, so r > 60.

Answer: (c) (a + b, c + d)

Method. diameter form: (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0

The equation is the circle on the diameter joining (2a, 2c) and (2b, 2d).
Centre = mid-point = (2a + 2b2, 2c + 2d2) = (a + b, c + d).

Answer: (d) (x − x₁) (x − x₂) + (y − y₁) (y − y₂) = 0

Method. diameter form of a circle: angle in a semicircle is a right angle

For any point P(x, y) on the circle, PA ⊥ PB where A(x₁, y₁), B(x₂, y₂) are the ends of the diameter.
Slope condition: y − y₁x − x₁ · y − y₂x − x₂ = -1.
Hence (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0.

Answer: (c) 4x² + 4y² - 142x + 47y + 138 = 0

Method. general circle x² + y² + 2gx + 2fy + c = 0 through three points

(1, 0): 1 + 2g + c = 0
(0, -6): 36 − 12f + c = 0
(3, 4): 25 + 6g + 8f + c = 0
From the first two, c = -1 − 2g and 12f = 35 − 2g; substituting in the third: 24 + 4g + 2(35 − 2g)3 = 0, so g = -714.
Then c = 692, f = 478.
x² + y² - 712x + 474y + 692 = 0; multiplying by 4: 4x² + 4y² - 142x + 47y + 138 = 0.

Shortcut. (0, -6) must satisfy it: 144 − 6(47) + 138 = 0 works only with +47y and +138, i.e. option (c).

Answer: (b) x² + y² + 3x + 12y + 2 = 0

Method. centre on the given line and equidistant from the two given points

Let the centre be (h, 2h − 3), since it lies on 2x − y − 3 = 0.
Equal distances to (3, -2) and (-2, 0):
(h − 3)² + (2h − 1)² = (h + 2)² + (2h − 3)²
5h² - 10h + 10 = 5h² - 8h + 13, so h = -32 and the centre is (-32, -6).
r² = (-2 + 32)² + (0 + 6)² = 1454.
(x + 32)² + (y + 6)² = 1454 gives x² + y² + 3x + 12y + 2 = 0.

Shortcut. Test the options: only (b) is satisfied by both (3, -2) and (-2, 0) [9 + 4 + 9 − 24 + 2 = 0; 4 − 6 + 2 = 0].

Answer: (a) (⁠√2,⁠ 0)

Method. centre lies on a bisector of the two lines; tangent length from the origin is 1

The circle touches both lines, so its centre lies on a bisector of x − y = 0 and x + y = 0, i.e. on the x-axis or the y-axis; take (h, 0).
Radius = distance to x − y = 0 = |h|⁠√2.⁠
The points of contact are at distance 1 from O, so the tangent length from O is 1: OC² = r² + 1.
h² = h²2 + 1, so h² = 2 and h = ±⁠√2.⁠
(⁠√2,⁠ 0) is a possible centre (check: foot of the perpendicular from (⁠√2,⁠ 0) to y = x is (1⁠√2,⁠ 1⁠√2),⁠ at distance 1 from O).

Answer: (a) ⁠√2⁠ k

Method. circle touching both axes: points of contact are (k, 0) and (0, k)

(x − k)² + (y − k)² = k²: centre (k, k), radius k (k > 0).
Put y = 0: (x − k)² = 0, so P = (k, 0); put x = 0: Q = (0, k).
PQ = ⁠√(k²⁠ + k²) = ⁠√2⁠ k.

Answer: (b) Only two

Method. read the centre and radius from the standard form

Divide by 4: x² + y² - ax − ay + a²4 = 0.
Centre (a2, a2); r² = a²4 + a²4 − a²4 = a²4, so r = |a|2.
1. Distance of the centre from each axis = |a|2 = r: touches both axes. Correct.
2. Diameter = 2r = |a|, not 2a. Wrong.
3. a2 + a2 = a: the centre lies on x + y = a. Correct.
Exactly two statements are correct.

Answer: (c) x² + y² - 2x − 2y + 1 = 0

Method. circle touching both axes in the first quadrant has centre (r, r)

Centre (r, r), radius r.
It touches y = 2, so the distance from the centre to that line equals r: |r − 2| = r.
r − 2 = r is impossible, so 2 − r = r and r = 1.
(x − 1)² + (y − 1)² = 1, i.e. x² + y² - 2x − 2y + 1 = 0.

Answer: (b) (-g, -f2)

Method. normal at a point of contact passes through the centre, so it meets the circle again at the other end of that diameter

Centre (-g2, -f2).
The circle touches the y-axis, so the point of contact is the foot of the perpendicular from the centre: (0, -f2).
The normal there is the radius through the centre, the line y = -f2.
It meets the circle again at the diametrically opposite point: 2(-g2) − 0 = -g, so (-g, -f2).

Answer: (d) ⁠√(2⁠ - ⁠√2)⁠ units

Method. chord = 2r sin(θ2)

r = 1, θ = 45°, so chord = 2 sin 22.5°.
sin² 22.5° = 1 − cos 45°2 = 2 - ⁠√24,⁠ so sin 22.5° = ⁠√(2⁠ - ⁠√2)⁠2.
Chord = 2 · ⁠√(2⁠ - ⁠√2)⁠2 = ⁠√(2⁠ - ⁠√2)⁠ units.

Shortcut. Cosine rule: chord² = 1 + 1 − 2 cos 45° = 2 - ⁠√2.⁠

Answer: (c) (-1 + 1⁠√2,⁠ -1 − 1⁠√2)⁠

Method. vertices of an inscribed square with sides parallel to the axes are at (h ± r⁠√2,⁠ k ± r⁠√2)⁠

(x + 1)² + (y + 1)² = 1: centre (-1, -1), r = 1.
The diagonals of the square are diameters; with sides parallel to the axes they lie along y + 1 = ±(x + 1).
So each vertex is at distance 1 from the centre along a 45° direction: (-1 ± 1⁠√2,⁠ -1 ± 1⁠√2).⁠
(-1 + 1⁠√2,⁠ -1 − 1⁠√2)⁠ is one of them.
(-2, 2) and (-2, -2) are at distances ⁠√10⁠ and ⁠√2⁠ from the centre, not on the circle.

Answer: (c) π/⁠4 or 3π/⁠4

Method. angle at the centre is twice the angle at the circumference

Centre O(0, 0). OB⃗ · OC⃗ = (6)(-8) + (8)(6) = 0, so ∠BOC = π2.
If A is on the major arc, ∠BAC = 12 ∠BOC = π4.
If A is on the minor arc, ∠BAC = π − π4 = 3π4 (opposite angles of a cyclic quadrilateral).
So ∠BAC = π4 or 3π4.

Answer: (d) Cannot be determined due to insufficient data

Method. the data fix only B, C and the circle; A can be any other point of the circle

A is only required to lie on x² + y² = 100 and differ from B and C.
(-6, 8), (-6, -8) and (5⁠√2,⁠ 5⁠√2)⁠ all satisfy x² + y² = 100 (36 + 64 = 100; 50 + 50 = 100), so each is a possible A.
Nothing in the data picks one of them: the coordinates of A cannot be determined.

Answer: (b) (32, 12)

Method. mid-point of a chord is the foot of the perpendicular from the centre to the chord

Centre (0, 0). The perpendicular from O to 3x + y = 5 is along (3, 1): points (3t, t).
3(3t) + t = 5 gives t = 12.
Mid-point = (32, 12).
Check: 3(32) + 12 − 5 = 0.

Answer: (b) x² + y² - ax − ay = 0

Method. find the chord's end points, then use the diameter form

Put y = a − x in x² + y² = a²: 2x² - 2ax = 0, so x = 0 or x = a.
End points of the chord: (a, 0) and (0, a).
Circle on this diameter: (x − a)(x − 0) + (y − 0)(y − a) = 0
x² + y² - ax − ay = 0.

Answer: (a) 2 < r < 8

Method. two circles cut in two points when |r₁ - r₂| < C₁C₂ < r₁ + r₂

Circle 1: centre (0, 0), radius r.
Circle 2: (x − 5)² + y² = 25 − 16 = 9, so centre (5, 0), radius 3.
Distance between centres = 5.
Two distinct points of intersection need |r − 3| < 5 < r + 3.
5 < r + 3 gives r > 2; |r − 3| < 5 gives -2 < r < 8.
Hence 2 < r < 8.