Q.The equation of the circle whose radius is 4 and centre is $(0, 1)$ is

Showing the 12 most recent of 21 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.Centre of the circle $x^2 + y^2 + 8x + 10y - 8 = 0$ is

   (a) $(8, 10)$

   (b) $(4, 5)$

   (c) $(-4, 5)$

   (d) $(-4, -5)$

   ›Reveal solutionSolution

   Match the given equation to the general circle form $x^2+y^2+2gx+2fy+c=0$ and read off the centre $(-g,-f)$.

   The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$, with centre $(-g,-f)$.

   …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.The coordinates of the centre of the circle $x^2 + y^2 = a^2$ is —

   (a) $(1, 1)$

   (b) $(1, 0)$

   (c) $(0, 0)$

   (d) $(0, 1)$

   ›Reveal solutionSolution

   The centre is $(0,0)$, option (c).

   The general equation of a circle with centre $(h,k)$ and radius $r$ is $(x-h)^2+(y-k)^2=r^2$.

   …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.The centre of the circle given by the equation x² + y² - 8x + 12y - 12 = 0 has the coordinates:

   (a) (8, -12)

   (b) (-8, 12)

   (c) (4, -6)

   (d) (-4, 6)

   ›Reveal solutionSolution

   Comparing with the general circle equation x²+y²+2gx+2fy+c=0, the centre is (−g, −f).

   The general form of a circle is:

   $x^2+y^2+2gx+2fy+c=0, \quad \text{centre} = (-g,-f)$

   Given $x^2+y^2-8x+12y-12=0$, compare coefficients: …

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4. CBSE 2025Set ANNUAL1 markMCQ
   Q.The diameter of the circle $x^2 + y^2 = 256$ is

   (a) 16

   (b) 32

   (c) 8

   (d) 64

   ›Reveal solutionSolution

   Diameter of $x^2+y^2=256$ is $32$.

   …

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5. CBSE 2025Set ANNUAL1 markMCQ
   Q.The centre of the circle $x^2 + y^2 = 9$ is

   (a) $(9, 0)$

   (b) $(0, 9)$

   (c) $(9, 9)$

   (d) $(0, 0)$

   ›Reveal solutionSolution

   Centre of $x^2+y^2=9$ is $(0,0)$.

   …

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6. CBSE 2025Set ANNUAL1 markMCQ
   Q.A circle passes through the points $A(2, -9)$, $B(5, -8)$ and $C(2, 1)$. The centre of the circle is

   (a) $(2, -4)$

   (b) $(-3, 4)$

   (c) $\left(3, \dfrac{-16}{3}\right)$

   (d) None of these

   ›Reveal solutionSolution

   Substituting all three points into $x^2+y^2+2gx+2fy+c=0$ and solving the resulting linear system gives centre $(-g,-f) = (2,-4)$.

   Let the circle be $x^2+y^2+2gx+2fy+c=0$.

   Substitute $A(2,-9)$: $4+81+4g-18f+c=0 \implies 4g-18f+c=-85$ ...(1)

   Substitute $B(5,-8)$: $25+64+10g-16f+c=0 \implies 10g-16f+c=-89$ ...(2)

   Substitute $C(2,1)$: $4+1+4g+2f+c=0 \implies 4g+2f+c=-5$ ...(3)

   …

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7. CBSE 2024Set ANNUAL1 markMCQ
   Q.Find the centre of the circle whose equation is $x^2+y^2-8x+10y-2=0$.

   (a) $(4,-5)$

   (b) $(-5,4)$

   (c) $(5,3)$

   (d) None of these

   ›Reveal solutionSolution

   Compare the given equation to the general circle form $x^2+y^2+Dx+Ey+F=0$, whose centre is $\left(-\dfrac D2, -\dfrac E2\right)$.

   The general equation of a circle is:

   $x^2 + y^2 + Dx + Ey + F = 0, \quad \text{centre} = \left(-\frac{D}{2}, -\frac{E}{2}\right)$

   …

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8. CBSE 2024Set ANNUAL1 markMCQ
   Q.Find the equation of the circle whose centre is $(2,2)$ and passes through the point $(4,5)$.

   (a) $x^2+y^2+4x+4y-5=0$

   (b) $x^2+y^2-4x-4y-5=0$

   (c) $x^2+y^2=27$

   (d) None of these

   ›Reveal solutionSolution

   Find the radius using the distance from centre to the given point, then write $(x-h)^2+(y-k)^2=r^2$ and expand.

   The centre is $(2,2)$ and the circle passes through $(4,5)$, so the radius is the distance between them:

   $r^2 = (4-2)^2 + (5-2)^2 = 2^2 + 3^2 = 4 + 9 = 13$

   The circle's equation is:

   $(x-2)^2 + (y-2)^2 = 13$

   …

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9. CBSE 2024Set ANNUAL1 mark
   Q.Find the equation of the circle with centre $(-3, 2)$ and radius 4 unit.

   ›Reveal solutionSolution

   The required circle's equation is $(x+3)^2+(y-2)^2=16$.

   The standard equation of a circle with centre $(h,k)$ and radius $r$ is $(x-h)^2+(y-k)^2=r^2$. Substituting the centre $(-3,2)$ and radius $4$: …

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10. CBSE 2024Set sz1 markMCQ
    Q.Equation of circle with centre $(0, 2)$ and radius $2$ is equal to:

    (a) $x^2 + y^2 - 6y = 0$

    (b) $x^2 + y^2 - 4y = 0$

    (c) $x^2 + 2y^2 - 3 = 0$

    (d) $2x^2 + y^2 + 3y = 0$

    ›Reveal solutionSolution

    The circle with centre $(0,2)$ and radius $2$ expands to $x^2+y^2-4y=0$.

    The standard equation of a circle with centre $(h,k)$ and radius $r$ is:

    $(x-h)^2 + (y-k)^2 = r^2$

    Here $h=0, k=2, r=2$:

    $(x-0)^2 + (y-2)^2 = 2^2$ …

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11. CBSE 2023Set ANNUAL1 markMCQ
    Q.The equation of the circle whose radius is 4 and centre is $(0, 1)$ is

    (a) $x^2+y^2=16$

    (b) $x^2+(y-1)^2=16$

    (c) $(x-1)^2+y^2=16$

    (d) $x^2+(y+1)^2=16$

    ›Reveal solutionSolution

    Plugging centre $(0,1)$ and radius $4$ into the standard circle equation gives $x^2+(y-1)^2=16$.

    The standard equation of a circle with centre $(h,k)$ and radius $r$ is:

    $(x-h)^2+(y-k)^2=r^2$

    …

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12. CBSE 2023Set ANNUAL1 markMCQ
    Q.The equation of the circle with centre $(1,1)$ and radius $\sqrt2$ units will be:

    (a) $x^2+y^2-2x-2y=0$

    (b) $x^2+y^2+2x+2y=0$

    (c) $(x+1)^2+(y+1)^2=2$

    (d) none of these

    ›Reveal solutionSolution

    The circle with centre $(1,1)$ and radius $\sqrt2$ has equation $x^2+y^2-2x-2y=0$.

    The standard equation of a circle with centre $(h,k)$ and radius $r$ is $(x-h)^2+(y-k)^2=r^2$. Here $h=1,k=1,r=\sqrt2$ so $r^2=2$:

    $(x-1)^2+(y-1)^2=2$

    Expanding:

    …

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