Q.Find the equation of the circle passing through the points $(2, 3)$ and $(-1, 1)$ and whose centre is on the line $x - 3y - 11 = 0$.

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:

Now square both sides to remove the square root:

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

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.

Quick Example

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

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

Simplify:

$(x + 1)^2 + (y - 4)^2 = 9$ …