Questions & explanations
1. A circle through the intersection of x²+y²=4 and x²+y²-2x-4=0 also passes through (2,3). Is it orthogonal to x²+y²-6x-8y+24=0?
- Yes
- Cannot be determined
- No
- Only if the circle is unique
Answer: No
Family: (x²+y²-4)+λ(x²+y²-2x-4)=0. Substitute (2,3): (4+9-4)+λ(4+9-4-4)=0 → 9+5λ=0 → λ=-9/5. Circle: (x²+y²-4) - (9/5)(x²+y²-2x-4)=0 → multiply 5: 5x²+5y²-20-9x²-9y²+18x+36=0 → -4x²-4y²+18x+16=0 → divide -2: 2x²+2y²-9x-8=0. General form: x²+y²-(9/2)x-4=0 → g=-9/4, f=0, c=-4. For S₃: x²+y²-6x-8y+24=0 → g'=-3, f'=-4, c'=24. Orthogonality: 2gg'+2ff' = c+c' → 2(-9/4)(-3)+0 = -4+24 → (27/2)=20 → false. So not orthogonal.
2. For the circle x² + y² = 25, find the locus of midpoints of chords that subtend a right angle at the centre.
- x² + y² = 25/2
- x² + y² = 25
- x² + y² = 25/4
- x² + y² = 25√2
Answer: x² + y² = 25/2
Let midpoint be (x₁, y₁). Chord equation is T = S₁: xx₁ + yy₁ = x₁² + y₁². Distance from centre to chord = √(x₁² + y₁²). Chord length = 2√(25 - (x₁² + y₁²)). For chord to subtend 90° at centre, chord length = r√2 = 5√2. Equating: 2√(25 - (x₁² + y₁²)) = 5√2 → 25 - (x₁² + y₁²) = 25/2 → x₁² + y₁² = 25/2. Hence locus is x² + y² = 25/2.
3. Given three circles S₁: x²+y²=1, S₂: x²+y²-2x=0, S₃: x²+y²-2y=0. Which of the following is true?
- The radical centre is (1, 1) and the circle x²+y²-2x-2y+1=0 cuts all three orthogonally.
- The radical centre is (1/2, 1/2) and no circle cuts all three orthogonally.
- The radical centre is (1/2, 1/2) and the circle x²+y²-2x-2y+1=0 cuts all three orthogonally.
- The radical centre is (1, 1) and no circle cuts all three orthogonally.
Answer: The radical centre is (1/2, 1/2) and no circle cuts all three orthogonally.
Radical axes: S₁-S₂ gives 2x=1 → x=1/2; S₁-S₃ gives 2y=1 → y=1/2; S₂-S₃ gives x=y. Intersection is (1/2,1/2). For a circle orthogonal to all three, solving conditions yields circle x²+y²+x+y-1=0, which is not among the options. Hence no circle in the options cuts all three orthogonally. Option b correctly states both facts.
4. The length of the tangent from point (7,1) to the circle x² + y² - 4x + 6y - 12 = 0 is:
- 6
- 5
- 4
- 3
Answer: 4
The length of the tangent from a point to a circle is √S₁, where S₁ is obtained by substituting the point into the circle equation. For circle x²+y²-4x+6y-12=0, center (2,-3), radius r=5. S₁ = 7²+1²-4·7+6·1-12 = 49+1-28+6-12 = 16. So tangent length = √16 = 4.
5. Two circles are given by S1: x² + y² - 4 = 0 and S2: x² + y² - 2x = 0. Which of the following is the equation of a circle passing through their intersection points?
- x² + y² - 2x - 4 = 0
- x² + y² - 2x + 4 = 0
- x² + y² - 2x - 2 = 0
- x² + y² - 2x + 2 = 0
Answer: x² + y² - 2x - 4 = 0
The family of circles through the intersection of S1 and S2 is S1 + λ S2 = 0, λ ≠ -1. Substituting S1 and S2 gives (x²+y²-4) + λ(x²+y²-2x)=0. For λ=1, we get 2x²+2y²-2x-4=0, which simplifies to x²+y²-2x-4=0. This circle passes through the intersection points.
6. For a circle with center (h,k) and radius r, which of the following is the parametric form?
- x = h + r secθ, y = k + r tanθ
- x = h + r sinθ, y = k + r cosθ
- x = h + r cosθ, y = k + r sinθ
- x = h + r tanθ, y = k + r secθ
Answer: x = h + r cosθ, y = k + r sinθ
The parametric form of a circle is derived from the definition of sine and cosine. For a point on the circle, the coordinates are (h + r cosθ, k + r sinθ) where θ is the angle measured from the positive x-direction. This satisfies (x-h)² + (y-k)² = r².
7. A circle through (1,1) and (2,2) is orthogonal to x² + y² = 4. Find its equation.
- x² + y² - 3x - 3y + 4 = 0
- x² + y² - 2x - 2y + 4 = 0
- x² + y² - 4x - 4y + 4 = 0
- x² + y² - 6x - 6y + 4 = 0
Answer: x² + y² - 3x - 3y + 4 = 0
Orthogonality with x²+y²=4 gives c=4. Let circle be x²+y²+2gx+2fy+4=0. Substituting (1,1): 2+2g+2f+4=0 => g+f=-3. Substituting (2,2): 8+4g+4f+4=0 => g+f=-3. So any circle with g+f=-3 works. Option a: g=-3/2, f=-3/2, sum=-3, satisfies.
8. Two circles are given by x² + y² - 6x + 4y + 9 = 0 and x² + y² - 2x - 4y - 3 = 0. Are they orthogonal?
- Yes, because 2(-3)(-1) + 2(2)(-2) = 6 - 8 = -2 equals 9 + (-3) = 6
- Yes, because 2(-3)(-1) + 2(2)(-2) = 6 + 8 = 14 equals 9 + (-3) = 6
- No, because 2(-3)(-1) + 2(2)(-2) = 6 - 8 = -2, which is not equal to 9 + (-3) = 6
- No, because 2(-3)(-1) + 2(2)(-2) = 6 + 8 = 14, which is not equal to 9 + (-3) = 6
Answer: No, because 2(-3)(-1) + 2(2)(-2) = 6 - 8 = -2, which is not equal to 9 + (-3) = 6
For orthogonal circles, condition is 2g₁g₂ + 2f₁f₂ = c₁ + c₂. Here g₁ = -3, f₁ = 2, c₁ = 9; g₂ = -1, f₂ = -2, c₂ = -3. Left side: 2(-3)(-1) + 2(2)(-2) = 6 - 8 = -2. Right side: 9 + (-3) = 6. Since -2 ≠ 6, circles are not orthogonal.
9. The director circle of the circle x² + y² = 9 is
- x² + y² = 9
- x² + y² = 18
- x² + y² = 3
- x² + y² = 81
Answer: x² + y² = 18
The director circle is the locus of points from which perpendicular tangents can be drawn. For a circle of radius r, the director circle is concentric with radius r√2. Here r = 3, so radius = 3√2, giving equation x² + y² = 18.
10. Which condition must a second-degree equation satisfy to represent a circle?
- A = C and B = 0
- A = C and B ≠ 0
- A = -C and B = 0
- A = C and B = 1
Answer: A = C and B = 0
For Ax² + Bxy + Cy² + Dx + Ey + F = 0 to represent a circle, the coefficients of x² and y² must be equal (A = C) and the xy term must be absent (B = 0). This ensures the equation can be reduced to the standard circle form.
11. Find the equation of the circle passing through (1,0), (0,1), and (-1,0).
- x² + y² + 2y = 1
- x² + y² + 2x = 1
- x² + y² - 2x = 1
- x² + y² = 1
Answer: x² + y² = 1
Using general circle equation x²+y²+2gx+2fy+c=0. Substituting (1,0): 1+2g+c=0. (0,1): 1+2f+c=0. (-1,0): 1-2g+c=0. Subtract first and third: 4g=0 → g=0. Then c=-1. From second: 1+2f-1=0 → f=0. Equation: x²+y²-1=0 → x²+y²=1.
12. A circle is defined as the set of all points in a plane that are at a fixed distance from a fixed point. If the fixed point is (h,k) and the fixed distance is r, what is the equation of the circle?
- (x-h)^2 + (y-k)^2 = r^2
- (x-h)^2 + (y-k)^2 = r
- (x+h)^2 + (y+k)^2 = r^2
- (x-h)^2 + (y-k)^2 = 2r
Answer: (x-h)^2 + (y-k)^2 = r^2
By definition, the distance from any point (x,y) to the center (h,k) equals r. Using the distance formula, sqrt((x-h)^2 + (y-k)^2) = r. Squaring both sides gives (x-h)^2 + (y-k)^2 = r^2. Hence option a is correct.