Find the equation of a circle with centre (-2, 3) and radius 4.

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(-2, 3)

Meghalaya MboseMBOSE Meghalaya 11th Board 2022Subjective· 1mImportance★★★★★

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Concept understanding — Circle Equation Standard Form

Where the Circle Equation Comes From

Imagine you're standing at a point on a flat field. You tie a rope to a stake at that point, walk out the full length of the rope, and start walking in a circle, keeping the rope taut. Every point you step on is exactly the same distance from the stake.

That's the entire idea: a circle is the set of all points that are a fixed distance (the radius) from a fixed point (the centre).

If we put this on a coordinate plane, we can turn that geometric idea into an algebraic equation.

From Geometry to Algebra

Let the centre be at coordinates $(h, k)$ . Let the radius be $r$ . Take any point $(x, y)$ that lies on the circle. The distance from $(x, y)$ to $(h, k)$ must equal $r$ .

What's the distance between two points in the plane? The distance formula:

\sqrt{(x - h)^2 + (y - k)^2} = r

Now square both sides to remove the square root:

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

That's it. That's the standard form of the equation of a circle.

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

What Each Piece Tells You

$(x - h)$ and $(y - k)$ — these shift the circle away from the origin. If the centre is at $(0,0)$ , the equation simplifies to $x^2 + y^2 = r^2$ .
$r^2$ — notice it's the square of the radius, not the radius itself. If the equation says $x^2 + y^2 = 25$ , the radius is $\sqrt{25} = 5$ , not 25.
The equals sign — only the points $(x, y)$ that make this equation true lie on the circle. Any other point gives a larger or smaller left-hand side.

Watch out

A common mistake: for $(x - 3)^2 + (y + 2)^2 = 16$ , students often read the centre straight off the signs printed in the equation and say $(3, 2)$ . That's wrong. Each bracket must first be written in the exact form $x - h$ and $y - k$ : here $(y + 2) = (y - (-2))$ , so $k = -2$ , not $2$ . The centre is actually $(3, -2)$ . Always flip the sign inside every bracket before reading off $h$ and $k$ .

Quick Example

Write the equation of a circle with centre $(-1, 4)$ and radius $3$ .

Here $h = -1$ , $k = 4$ , $r = 3$ . Plug in:

(x - (-1))^2 + (y - 4)^2 = 3^2

Simplify:

$(x + 1)^2 + (y - 4)^2 = 9$ …

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$(x+2)^2+(y-3)^2=16$ .

Standard form: $(x-h)^2+(y-k)^2=r^2$ with centre $(h,k)=(-2,3)$ and $r=4$ :

$(x-(-2))^2+(y-3)^2 = 4^2$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 21 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$x^2 + y^2 + 8x + 10y - 8 = 0$
(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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CBSE 2026Set ANNUAL1 markMCQ
$x^2 + y^2 = a^2$
(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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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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CBSE 2025Set ANNUAL1 markMCQ
$x^2 + y^2 = 256$
(a) 16
(b) 32
(c) 8
(d) 64
›Reveal solutionSolution

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

…

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CBSE 2025Set ANNUAL1 markMCQ
$x^2 + y^2 = 9$
(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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CBSE 2025Set ANNUAL1 markMCQ
$A(2, -9)$
(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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CBSE 2024Set ANNUAL1 markMCQ
$x^2+y^2-8x+10y-2=0$
(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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CBSE 2024Set ANNUAL1 markMCQ
$(2,2)$
(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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CBSE 2024Set ANNUAL1 mark
$(-3, 2)$
›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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CBSE 2024Set sz1 markMCQ
$(0, 2)$
(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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CBSE 2023Set ANNUAL1 markMCQ
$(0, 1)$
(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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CBSE 2023Set ANNUAL1 markMCQ
$(1,1)$
(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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←Previous chapterCh 9Straight Lines

Q.Find the equation of a circle with centre (−2,3)(-2, 3)(−2,3) and radius 4.

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Q.Find the equation of a circle with centre $(-2, 3)$ and radius 4.

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