Pair of Tangents from a Point: Formula and Examples

The pair of tangents from an external point (x₁, y₁) to the circle S = x² + y² + 2gx + 2fy + c = 0 is captured by one combined equation: SS₁ = T², where S is the circle's left side, S₁ is S evaluated at the point, and T is the T = 0 expression. Expanding it gives a second-degree equation that factorises into the two tangent lines.

This matters because finding tangent slopes one by one is slow and error-prone, while SS₁ = T² writes both tangents at once. The formula also doubles as a position test: SS₁ − T² keeps its sign structure so that the equation degenerates correctly when the point lies on the circle (giving the single tangent, counted twice). It is a standard result asked directly in board and entrance exams.

Key Concepts

• Combined equation: SS₁ = T², with S = x² + y² + 2gx + 2fy + c, S₁ its value at (x₁, y₁), and T = xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c.
• Factorisation: the expanded equation is a homogeneous-style second-degree equation that splits into two linear factors — the two tangents.
• On-circle case: if S₁ = 0, the equation becomes T² = 0, i.e. the tangent at the point, doubled.
• Director circle note: the locus of points from which tangents are perpendicular is the director circle x² + y² = 2r² for S = x² + y² − r².

Formulas

• Pair of tangents: (x² + y² + 2gx + 2fy + c)(x₁² + y₁² + 2gx₁ + 2fy₁ + c) = [xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c]².
• Units: every term is a squared area, so the equation is dimensionally consistent.
• Assumption: S₁ > 0 (external point) for two distinct real tangents; S₁ < 0 gives an equation with no real lines.
• The result is always a pair of straight lines through (x₁, y₁) — substitute the point to verify.

Worked Example

Find the pair of tangents from (7, 1) to the circle x² + y² = 25.

1. Compute S = x² + y² − 25, S₁ = 49 + 1 − 25 = 25, T = 7x + y − 25.
2. Write SS₁ = T²: 25(x² + y² − 25) = (7x + y − 25)².
3. Expand the right side: 49x² + y² + 625 + 14xy − 350x − 50y.
4. Bring all left: 0 = 24x² − 24y² + 14xy − 350x − 50y + 1250; divide by 2: 12x² − 12y² + 7xy − 175x − 25y + 625 = 0.
5. Factorise: (3x + 4y − 25)(4x − 3y − 25) = 0 — check by expanding: 12x² − 9xy − 75x + 16xy − 12y² − 100y − 100x + 75y + 625 matches exactly.
6. The tangents are 3x + 4y = 25 and 4x − 3y = 25, both passing through (7, 1): 21 + 4 = 25 and 28 − 3 = 25. Correct.

Common Misconceptions

• "SS₁ = T² gives one tangent." It gives both tangents combined; you must factorise (or solve) to separate them.
• "It works for interior points too." For S₁ < 0 the equation represents no real lines — geometrically there are no real tangents.
• "The expanded equation is a circle." The x² and y² coefficients differ in general, so it is a pair of lines, not a circle.

Real-World Applications

• Visibility polygons: in games and robotics, the two tangents from an observer to a circular obstacle bound the shadow region behind it.
• Camera framing: the tangents from a camera position to a circular object define the extreme rays that just include it.
• Satellite coverage: tangent lines from a satellite to Earth's disc mark the horizon limits of its footprint.
• CNC machining: tool paths that graze a circular pocket are computed as tangent pairs from the tool's start point.

In short: SS₁ = T² packages both tangents from an external point into one equation — expand, factorise, and read off the two lines. Check S₁ > 0 first, verify both factors pass through the external point, and remember the on-circle special case collapses to T² = 0.