It's assumed that all points are co-planar, so nothing fancy going on here. Refer to the illustration above. Clicking it should return a larger (readable) version.
Given:
• a circle, (x - h)² + (y - k)² = r²,
with center at (h, k) and radius r; and
• a point at (p, q) outside the circle:
Find:
• the coordinates of the tangent point(s), and
• the equation(s) of the tangent line(s) from the point to the circle.
Through any given outside point, there are two possible tangent lines to a given circle, but once you find one, finding its doppelgängers is just a matter of changing a few signs.
Let Δ = ±1 to express that sign change.
Let ∆ABC be such that:
• A is the vertex at (h, k), which is the center of the given circle,
• B is the vertex at given point at (p, q), and
• C is the tangent point.
Notice that ∆ABC is a right triangle by definition because any tangent line is perpendicular to the radius at the point of intersection.
So:
c = AB (the hypotenuse as shown) = √([p - h)² + (q - k)²]α = ∠CAB = acos[r / c]
β = ∠ABC = asin[r / c] — reference only, not used in these calculations
γ = ∠BCA = π/2 — reference only, not used in these calculations
δ = acos[(p - h) / c]
(if you are wondering, α is "alpha," β is "beta," γ is "gamma," δ is small "delta," and Δ is capital "delta.")
δ is just used for the angle between the x-axis and c.
(See: Convert cartesian ↔ polar coordinates for some more details if you like.)
Tangent Point CΔ:
Δ = ±1
x = r · cos[α + Δδ] + h, and
y = Δr · sin[α + Δδ] + k
Tangent Line, y = mΔx + bΔ:
Δ = ±1
mΔ = –Δ/tan[α + Δδ], and
bΔ = Δ(r · cos[α + Δδ] + h) / tan[α + Δδ] + Δr · sin[α + Δδ] + k
Sample Calculation
Given:
Circle: (x - 0.6)² + (y - 0.5)² = 1.5625
Point: (-0.625, 2.25)
h = 0.6; k = 0.5; r = 1.25
p = -0.625; q = 2.25
δ ≈ 2.18152
α ≈ 0.94571
Δ=1:Tangent Point: (-0.6499, 0.5179); Tangent Line: y = 69.641x +45.7756
Δ=-1:
Tangent Point: (1.0109, 1.6805); Tangent Line: y = -0.3481x +2.0324
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