Question types

The Circle question types

116 questions across 6 question groups — pick any mix to generate a Maths paper with step-by-step answer keys.

116
Questions
6
Question groups
5
Question types
Sample Questions

The Circle questions

One sample from each question group in this chapter. Select any group above to see the full set with answer keys.

Q 1MCQ1 Mark
The equation of the incircle formed by the coordinate axes and the line $4x + 3y = 6$ is:
  • A
    $x^2 + y^2 - 6x - 6y + 9 = 0$
  • $4 (x^2 + y^2 - x - y) + 1 = 0$
  • C
    $4 (x^2 + y^2 + x + y) + 1 = 0$
  • D
    None of these

Answer: B.

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Q 2MCQ1 Mark
If the centroid of an equilateral triangle is $(1, 1)$ and its one vertex is $(-1, 2),$ then the equation of its circumcircle is:
  • $x^2 + y^2 - 2x - 2y - 3 = 0$
  • B
    $x^2 + y^2 + 2x - 2y - 3 = 0$
  • C
    $x^2 + y^2 + 2x + 2y - 3 = 0$
  • D
    None of these

Answer: A.

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Q 3MCQ1 Mark
The equation of the circle which touches the axes of coordinates and the line $\frac{\text{x}}{3}+\frac{\text{y}}{4}=1$ and whose centres lie in the first quadrant is $x^2 + y^2 − 2cx − 2cy + c^2 = 0,$ where $c$ is equal to:
  • A
    $4$
  • B
    $2$
  • C
    $3$
  • $6$

Answer: D.

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Q 4MCQ1 Mark
The equation of the circle passing through the point $(1, 1)$ and having two diameters along the pair of lines $x^2 - y^2 - 2x + 4y - 3 = 0,$ is:
  • $x^2 + y^2 - 2x - 4y + 4 = 0$
  • B
    $x^2 + y^2 + 2x + 4y - 4 = 0$
  • C
    $x^2 + y^2 - 2x + 4y + 4 = 0$
  • D
    None of these

Answer: A.

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Q 5MCQ1 Mark
If the point $(2, k)$ lies outside the circles $x^2 + y^2 + x - 2y - 14 = 0$ and $x^2 + y^2 = 13$ then klies in the interval:
  • A
    $(-3,\ -2)\cup(3,\ 4)$
  • B
    $-3,\ 4$
  • $(-\infty,\ -3)\cup(4,\ \infty)$
  • D
    $(-\infty,\ -2)\cup(3,\ \infty)$

Answer: C.

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Show that the point $(\text{x},\ \text{y})$ given by $\text{x}=\frac{2\text{at}}{1+\text{t}^2}$ and $\text{y}=\text{a}\Big(\frac{1-\text{t}^2}{1+\text{t}^2}\Big)$2 lies on a circle for all real values of t such that $-1\leq\text{t}\leq1,$ where a is any given real number.
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Show that the point $(\text{x},\ \text{y})$ given by $\text{x}=\frac{2\text{at}}{1+\text{t}^2}$ and $\text{y}=\text{a}\Big(\frac{1-\text{t}^2}{1+\text{t}^2}\Big)$2 lies on a circle for all real values of t such that $-1\leq\text{t}\leq1,$ where a is any given real number.
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