A circle is the locus (set) of all points in a plane that are equidistant (a constant distance) from a fixed point.
This is the most direct translation of the definition.
If the center is C(h, k) and the radius is r, then by the distance formula, any point P(x, y) on the circle must satisfy:
sqrt( (x - h)² + (y - k)² ) = r
Squaring both sides gives the standard form:
(x - h)² + (y - k)² = r²
This is a special case of the Center-Radius form where the center is at the origin (0, 0).
x² + y² = r²
This form is used when the endpoints of a diameter are known, say A(x₁, y₁) and B(x₂, y₂).
A key property of circles is that the angle subtended by a diameter at any point on the circumference is 90°.
So, if P(x, y) is any point on the circle, the line segment PA is perpendicular to the line segment PB.
(Slope of PA) × (Slope of PB) = -1
[ (y - y₁) / (x - x₁) ] · [ (y - y₂) / (x - x₂) ] = -1
(y - y₁)(y - y₂) = -(x - x₁)(x - x₂)
(x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0
Any three non-collinear points uniquely define a circle. To find its equation:
Let the three points be (x₁, y₁), (x₂, y₂), and (x₃, y₃).
Let the circle's equation be the general form x² + y² + 2gx + 2fy + c = 0 (see next section).
x₁² + y₁² + 2gx₁ + 2fy₁ + c = 0x₂² + y₂² + 2gx₂ + 2fy₂ + c = 0x₃² + y₃² + 2gx₃ + 2fy₃ + c = 0You now have a system of three linear equations for the three unknowns (g, f, and c). Solve this system to find the values of g, f, and c, then plug them back into the general equation.
If we expand the Center-Radius form (x - h)² + (y - k)² = r²:
(x² - 2hx + h²) + (y² - 2ky + k²) = r²
x² + y² - 2hx - 2ky + (h² + k² - r²) = 0
This is of the form:
x² + y² + 2gx + 2fy + c = 0
This is the general equation of a circle.
-2h = 2g → h = -g-2k = 2f → k = -fc = h² + k² - r² → r² = h² + k² - c = (-g)² + (-f)² - cFor a circle given in general form:
(-g, -f)sqrt(g² + f² - c)To find the center and radius from x² + y² + 2gx + 2fy + c = 0:
sqrt(g² + f² - c)Note:
g² + f² - c > 0, it is a real circle.g² + f² - c = 0, it is a point circle (radius is 0).g² + f² - c < 0, it is an imaginary circle (no real locus).The full general equation of a 2nd degree conic is ax² + 2hxy + by² + 2gx + 2fy + c = 0.
For this to represent a circle, it must match the form x² + y² + 2gx + 2fy + c = 0 (or a multiple of it).
The two necessary conditions are:
a = b
h = 0
A tangent is a line that touches the circle at exactly one point, the "point of tangency".
A normal is a line perpendicular to the tangent at the point of tangency. For a circle, the normal always passes through the center.
The equation of the tangent to a circle at a point (x₁, y₁) on the circle can be found using the "T = 0" rule. We replace:
x² → x·x₁y² → y·y₁2x → x + x₁2y → y + y₁c → cFor circle x² + y² = r²:
x·x₁ + y·y₁ = r²For circle x² + y² + 2gx + 2fy + c = 0:
x·x₁ + y·y₁ + g(x + x₁) + f(y + y₁) + c = 0
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0The normal passes through (x₁, y₁) and the center (-g, -f). We can just use the two-point form of a line.
Slope of normal m = (y₁ - (-f)) / (x₁ - (-g)) = (y₁ + f) / (x₁ + g).
Using the point-slope form Y - y₁ = m(X - x₁):
Y - y₁ = [ (y₁ + f) / (x₁ + g) ] · (X - x₁)
A simpler form is often (Y - y₁)(x₁ + g) = (X - x₁)(y₁ + f).
For the simple circle x² + y² = r²:
The center is (0, 0). The slope of the normal is (y₁ - 0) / (x₁ - 0) = y₁/x₁.
Equation: Y - y₁ = (y₁/x₁) (X - x₁)
x₁(Y - y₁) = y₁(X - x₁) → x₁Y - x₁y₁ = y₁X - x₁y₁ → x₁Y = y₁X or y = (y₁/x₁)x.
x² + y² + 2gx + 2fy + c = 0.
(-g, -f). (Just halve the coefficients of x and y, and flip the signs).sqrt(g² + f² - c).Coeff(x²) = Coeff(y²). 2. Coeff(xy) = 0.(x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0. Very fast if you are given two endpoints.T = 0 replacement rule.
x² → xx₁2x → (x + x₁)x² + y² + 4x = 0 → xx₁ + yy₁ + 2(x + x₁) = 0.
(-g, -f) and the point (x₁, y₁) and write the equation of the line passing through them.