How this instrument works
A circle's standard equation, (x − h)² + (y − k)² = r², is really just the distance formula in another shape: it states that every point (x, y) on the circle sits exactly a distance r from the center (h, k). This calculator turns that equality into a test by computing (px − h)² + (py − k)² − r² for any point (px, py) — the result is zero exactly when the point lies ON the circle, negative when the point is INSIDE it, and positive when the point is OUTSIDE it.
This works because (px − h)² + (py − k)² is simply the squared distance from the test point to the center, and the test compares that squared distance to r², the circle's own squared radius. A point closer to the center than r has a smaller squared distance, giving a negative result; a point farther away gives a positive one — the sign alone answers the inside-or-outside question without ever needing to take a square root.
This is the same underlying equation used to convert between a circle's general algebraic form (with an x² + y² term expanded out) and its standard center-radius form — recognizing (x−h)² + (y−k)² = r² as a circle's signature, rather than treating it as an arbitrary equation, is the key skill behind both conversions.
- Enter the circle's center coordinates into the Center: x (h) and Center: y (k) fields.
- Enter the circle's radius into the Radius field.
- Enter the point you want to test into the Test point: x and Test point: y fields.
- Read the test value: exactly 0 means the point sits on the circle; negative means inside; positive means outside.
Worked example — testing (3, 4) against a radius-5 circle
A circle is centered at the origin (0, 0) with radius 5. Testing the point (3, 4): (3−0)² + (4−0)² − 5² = 9 + 16 − 25 = 0 — the point sits exactly on the circle's boundary, confirming the familiar 3-4-5 right triangle relationship in disguise.
Testing the center point itself, (0, 0), against that same circle: (0−0)² + (0−0)² − 25 = −25, a clearly negative result — the center is always as far inside the circle as possible, the most negative test value any point in that circle can produce.
Questions
What is the standard equation of a circle?
(x − h)² + (y − k)² = r², where (h, k) is the circle's center and r is its radius. It states that every point on the circle sits exactly a distance r from the center, which is exactly the distance formula rearranged.
How do I know if a point is inside or outside a circle?
Compute (px − h)² + (py − k)² − r² for the point. A result of exactly 0 means the point is ON the circle; a negative result means it's INSIDE; a positive result means it's OUTSIDE — the sign alone answers the question without needing a square root.
Why does the formula use squared distances instead of plain distance?
Comparing squared distances avoids an unnecessary square root and keeps the test purely algebraic — since squaring preserves the order of non-negative distances, comparing the squares gives exactly the same inside/outside/on answer as comparing the actual distances would.
How is this related to a circle's general form equation?
Expanding (x − h)² + (y − k)² = r² produces the general form x² + y² + Dx + Ey + F = 0, with D, E, and F absorbing h, k, and r. Converting between the two forms is a matter of completing the square in each variable.
What happens if the radius is zero?
The 'circle' collapses to a single point at (h, k), and the only point that tests to exactly 0 is that center point itself — every other point in the plane gives a strictly positive result, since a zero-radius circle contains nothing but its own center.