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Worked Examples · Example 1

Q.Find an equation of the circle with centre at (0,0)(0, 0) and radius rr.

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The equation of a circle centred at the origin with radius rr is x2+y2=r2x^2 + y^2 = r^2, derived directly from the distance formula — every point (x,y)(x, y) on the circle is exactly rr units from (0,0)(0, 0).

Why this works: the circle as a distance condition

A circle is the set of all points that are a fixed distance (the radius) from a fixed point (the centre). Here the centre is the origin (0,0)(0, 0) and the radius is rr. So the question becomes: which points (x,y)(x, y) are exactly rr units away from (0,0)(0, 0)?

The distance between any two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by the distance formula:

(x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

If we set (x1,y1)=(0,0)(x_1, y_1) = (0, 0) and (x2,y2)=(x,y)(x_2, y_2) = (x, y), the distance from the origin to (x,y)(x, y) is x2+y2\sqrt{x^2 + y^2}. For (x,y)(x, y) to lie on the circle, this distance must equal rr.

x2+y2=r\sqrt{x^2 + y^2} = r

Squaring both sides removes the square root and gives the clean, standard form.

Step-by-step derivation

  1. Write the distance condition. A point (x,y)(x, y) is on the circle if its distance from (0,0)(0, 0) is exactly rr:

(x−0)2+(y−0)2=r\sqrt{(x - 0)^2 + (y - 0)^2} = r

  1. Simplify inside the square root. (x−0)2=x2(x - 0)^2 = x^2 and (y−0)2=y2(y - 0)^2 = y^2, so:

x2+y2=r\sqrt{x^2 + y^2} = r

  1. Square both sides. This eliminates the square root. Since r≥0r \ge 0 (a radius is non-negative), squaring is safe:

x2+y2=r2x^2 + y^2 = r^2

That’s it — the equation of the circle.

Watch out

A common mistake is to forget the square on rr and write x2+y2=rx^2 + y^2 = r. Remember: the distance formula gives x2+y2\sqrt{x^2 + y^2}, and squaring that yields r2r^2, not rr.

What this equation tells you

  • Every pair (x,y)(x, y) that satisfies x2+y2=r2x^2 + y^2 = r^2 lies on the circle.
  • The circle is symmetric about both axes — replacing xx with −x-x or yy with −y-y leaves the equation unchanged.
  • If r=0r = 0, the equation becomes x2+y2=0x^2 + y^2 = 0, which only the point (0,0)(0, 0) satisfies — a degenerate circle (a single point).
Tip

This form is the simplest case of the general circle equation (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) is the centre. When h=k=0h = k = 0, you get x2+y2=r2x^2 + y^2 = r^2. Memorising the general form lets you handle any centre instantly.

✓Final answer

The equation of the circle is x2+y2=r2\boxed{x^2 + y^2 = r^2}.

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