Equation of a Circle

Statement

The equation of a circle with centre at the origin \((0,0)\) and radius \(r\) is:

\[ x^2 + y^2 = r^2 \]

This represents all points \((x,y)\) that are exactly distance \(r\) from the origin.

Why it’s true

• The distance from the origin to any point \((x,y)\) is \(\sqrt{x^2+y^2}\).
• On the circle, this distance is always equal to the radius \(r\).
• So, \(\sqrt{x^2+y^2}=r\). Squaring both sides gives \(x^2+y^2=r^2\).

Recipe (how to use it)

1. Square the x-coordinate and the y-coordinate of the point.
2. Add them together.
3. If the sum equals \(r^2\), the point lies on the circle.
4. If the sum is less than \(r^2\), the point is inside the circle.
5. If the sum is greater than \(r^2\), the point is outside the circle.

Spotting it

This formula is used whenever the circle is centred at the origin and you need to check if a point lies on it or describe its boundary.

Common pairings

• Distance formula (since it derives from it).
• Coordinate geometry problems involving circle equations.
• General circle equation: \((x-h)^2+(y-k)^2=r^2\) when the centre is \((h,k)\).

Mini examples

1. Circle: \(x^2+y^2=25\). Point: \((3,4)\). Check: \(3^2+4^2=9+16=25\). Lies on circle.
2. Circle: \(x^2+y^2=100\). Point: \((6,8)\). Check: \(36+64=100\). Lies on circle.
3. Circle: \(x^2+y^2=16\). Point: \((2,1)\). Check: \(4+1=5<16\). Point inside circle.

Pitfalls

• Forgetting the centre is at the origin (the general equation includes \((h,k)\)).
• Mixing up radius \(r\) with \(r^2\).

Exam strategy

• When checking if a point lies on a circle, always substitute into \(x^2+y^2\).
• Be careful with negative values: squaring ensures they contribute positively.

Summary

The equation \(x^2+y^2=r^2\) describes a circle of radius \(r\) centred at the origin. It comes directly from the distance formula and is a key result in coordinate geometry.