The area of the circle centred at $(1, 2)$ and passing through $(4, 6)$ is (A) $5\pi$ (B) $10\pi$ (C) $25\pi$ (D) none of these

The area of the circle centred at (1, 2) and passing through (4, 6) is (A) 5π (B) 10π (C) 25π (D) none of thes

NCERT Exemplar · Q33

(1, 2)

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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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The radius is the distance between the centre and the given point, found using the distance formula. That radius squared is $25$ , so the area is $25\pi$ . The correct option is (C).

The key idea: the area of a circle depends only on its radius. Here, the centre is fixed at $(1, 2)$ , and the circle passes through $(4, 6)$ . That means the distance from the centre to that point is exactly the radius. Once we have the radius, area follows directly from $\pi r^2$ .

The standard form of a circle’s equation is $(x - h)^2 + (y - k)^2 = r^2$ , where $(h, k)$ is the centre and $r$ is the radius. But we don’t need the full equation — just the radius.

Find the radius using the distance formula. The distance between $(1, 2)$ and $(4, 6)$ is:

r = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.

So the radius is $5$ units.

Compute the area. Area of a circle is $A = \pi r^2$ . Substituting $r = 5$ :

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Q.The area of the circle centred at (1,2)(1, 2)(1,2) and passing through (4,6)(4, 6)(4,6) is (A) 5π5\pi5π (B) 10π10\pi10π (C) 25π25\pi25π (D) none of these

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Q.The area of the circle centred at $(1, 2)$ and passing through $(4, 6)$ is
(A) $5\pi$
(B) $10\pi$
(C) $25\pi$
(D) none of these

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