If a circle passes through the point (0, 0), (a, 0), (0, b) then find the coordinates of its centre.

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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

That's the standard form. From this, you can immediately read off the centre $(-1, 4)$ and radius $3$ .

Why This Form Matters

The standard form is the most useful because it gives you the centre and radius at a glance. In exams, you'll often be given an expanded form like $x^2 + y^2 - 6x + 4y - 12 = 0$ and asked to rewrite it in standard form by completing the square — that's the next step in your learning, but the standard form itself is the destination.

For now: centre tells you where, radius tells you how big, and the equation tells you which points belong.

The Standard Form of a Circle's Equation is one of the first results in the NCERT Class 11 Mathematics chapter on Conic Sections, matching searches like "equation of a circle: definition, formula and examples" or "conic sections important questions class 11 maths". Recognising centre and radius directly from this form is also a routine, quick-scoring question type in CBSE boards, JEE Main, and state CET coordinate geometry sections.

The centre of the circle passing through $(0,0)$ , $(a,0)$ , and $(0,b)$ is $\left(\frac{a}{2}, \frac{b}{2}\right)$ . This follows because the perpendicular bisectors of the chords meet at the centre, and the given points form a right triangle whose hypotenuse is the diameter.

The three points given are $(0,0)$ , $(a,0)$ , and $(0,b)$ . Notice that $(a,0)$ lies on the x-axis and $(0,b)$ lies on the y-axis. So these three points form a right triangle with the right angle at the origin $(0,0)$ .

When a circle passes through all three vertices of a right triangle, the hypotenuse of that triangle is the diameter of the circle. This is a direct consequence of Thales' theorem: the angle subtended by a diameter at the circumference is a right angle.

Here, the hypotenuse is the line joining $(a,0)$ and $(0,b)$ . The centre of the circle is the midpoint of this diameter.

Identify the hypotenuse

The right angle is at $(0,0)$ , so the side opposite it — the line from $(a,0)$ to $(0,b)$ — is the hypotenuse.
Find the midpoint of the hypotenuse

The midpoint formula:

\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)

For $(a,0)$ and $(0,b)$ :

\text{Centre} = \left( \frac{a + 0}{2}, \frac{0 + b}{2} \right) = \left( \frac{a}{2}, \frac{b}{2} \right)

Verify using the general circle equation (optional but solidifying) The general equation of a circle is:

x^2 + y^2 + 2gx + 2fy + c = 0

Centre is $(-g, -f)$ .

Substituting $(0,0)$ gives $c = 0$ .

Substituting $(a,0)$ : $a^2 + 2ga = 0 \Rightarrow g = -\frac{a}{2}$ .

Substituting $(0,b)$ : $b^2 + 2fb = 0 \Rightarrow f = -\frac{b}{2}$ .

So centre $= (-g, -f) = \left( \frac{a}{2}, \frac{b}{2} \right)$ .

Watch out

A common mistake is to assume the centre is the midpoint of any two points. Only the midpoint of the hypotenuse works here because the right angle forces the hypotenuse to be the diameter.

Tip

Whenever you see three points where two lie on the axes and one is the origin, check if they form a right triangle. If yes, the circle's centre is simply the midpoint of the non-origin pair.

✓Final answer

\boxed{\left( \frac{a}{2}, \frac{b}{2} \right)}

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Q.If a circle passes through the point (0,0)(0, 0)(0,0), (a,0)(a, 0)(a,0), (0,b)(0, b)(0,b) then find the coordinates of its centre.

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Q.If a circle passes through the point $(0, 0)$ , $(a, 0)$ , $(0, b)$ then find the coordinates of its centre.

Unlock everything free for 14 days