Tangent lines to circles

Suppose that the equation of the circle in Cartesian coordinates is ${\displaystyle (x-a)^{2}+(y-b)^{2}=r^{2}}$ with center at (a, b). Then the tangent line of the circle at (x₁, y₁) has Cartesian equation

$${\displaystyle (x-x_{1})(x_{1}-a)+(y-y_{1})(y_{1}-b)=0}$$

This can be proved by taking the implicit derivative of the circle. Say that the circle has equation of ${\displaystyle (x-a)^{2}+(y-b)^{2}=r^{2},}$ and we are finding the slope of tangent line at (x₁, y₁) where ${\displaystyle (x_{1}-a)^{2}+(y_{1}-b)^{2}=r^{2}.}$ We begin by taking the implicit derivative with respect to x:

$${\displaystyle {\begin{aligned}{\overset {}{(}}x-a)^{2}+(y-b)^{2}&=r^{2}\\[4pt]2(x-a)+2(y-b){\frac {dy}{dx}}&=0\\[2pt]{\frac {dy}{dx}}&=-{\frac {x_{1}-a}{y_{1}-b}}\end{aligned}}}$$

Now that we have the slope of the tangent line, we can substitute the slope and the coordinate of the tangency point into the line equation y = kx + m.

$${\displaystyle {\begin{aligned}y&=-{\frac {x_{1}-a}{y_{1}-b}}x+y_{1}+x_{1}{\frac {x_{1}-a}{y_{1}-b}}\\y-y_{1}&=(x_{1}-x){\frac {x_{1}-a}{y_{1}-b}}\\[2pt](y-y_{1})(y_{1}-b)&=-(x-x_{1})(x_{1}-a)\\[8pt](x-x_{1})(x_{1}-a)+(y-y_{1})(y_{1}-b)&=0\end{aligned}}}$$

It is relatively straightforward to construct a line t tangent to a circle at a point T on the circumference of the circle:

• A line a is drawn from O, the center of the circle, through the radial point T;
• The line t is the perpendicular line to a.

Construction of a tangent to a given circle (black) from a given exterior point (P).

Thales' theorem may be used to construct the tangent lines to a point P external to the circle C:

• A circle is drawn centered on the midpoint of the line segment OP, having diameter OP, where O is again the center of the circle C.
• The intersection points T₁ and T₂ of the circle C and the new circle are the tangent points for lines passing through P, by the following argument.

The line segments OT₁ and OT₂ are radii of the circle C; since both are inscribed in a semicircle, they are perpendicular to the line segments PT₁ and PT₂, respectively. But only a tangent line is perpendicular to the radial line. Hence, the two lines from P and passing through T₁ and T₂ are tangent to the circle C.

Another method to construct the tangent lines to a point P external to the circle using only a straightedge:

• Draw any three different lines through the given point P that intersect the circle twice.
• Let A₁, A₂, B₁, B₂, C₁, C₂ be the six intersection points, with the same letter corresponding to the same line and the index 1 corresponding to the point closer to P.
• Let D be the point where the lines A₁B₂ and A₂B₁ intersect,
• Similarly E for the lines B₁C₂ and B₂C₁.
• Draw a line through D and E.
• This line meets the circle at two points, F and G.
• The tangents are the lines PF and PG.[1]

Let ${\displaystyle P=(a,b)}$ be a point of the circle with equation ${\displaystyle x^{2}+y^{2}=r^{2}.}$ The tangent at P has equation ${\displaystyle ax+by=r^{2},}$ because P lies on both the curves and ${\displaystyle {\vec {OP}}=(a,b)^{T}}$ is a normal vector of the line. The tangent intersects the x-axis at point ${\displaystyle P_{0}=(x_{0},0)}$ with ${\displaystyle ax_{0}=r^{2}.}$

Tangents through a point

Conversely, if one starts with point ${\displaystyle P_{0}=(x_{0},0),}$ than the two tangents through P₀ meet the circle at the two points ${\displaystyle P_{1/2}=(a,b_{\pm })}$ with

$${\displaystyle a={\frac {r^{2}}{x_{0}}},\qquad b_{\pm }=\pm {\sqrt {r^{2}-a^{2}}}=\pm {\frac {r}{x_{0}}}{\sqrt {x_{0}^{2}-r^{2}}}.}$$

Written in vector form:

$${\displaystyle {\binom {a}{b_{\pm }}}={\frac {r^{2}}{x_{0}}}{\binom {1}{0}}\pm {\frac {r}{x_{0}}}{\sqrt {x_{0}^{2}-r^{2}}}{\binom {0}{1}}\ .}$$

If point ${\displaystyle P_{0}=(x_{0},y_{0})}$ lies not on the x-axis: In the vector form one replaces x₀ by the distance ${\textstyle d_{0}={\sqrt {x_{0}^{2}+y_{0}^{2}}}}$ and the unit base vectors by the orthogonal unit vectors ${\textstyle {\vec {e}}_{1}={\frac {1}{d_{0}}}{\binom {x_{0}}{y_{0}}},\ {\vec {e}}_{2}={\frac {1}{d_{0}}}{\binom {-y_{0}}{x_{0}}}.}$ Then the tangents through point P₀ touch the circle at the points

$${\displaystyle {\binom {x_{1/2}}{y_{1/2}}}={\frac {r^{2}}{d_{0}^{2}}}{\binom {x_{0}}{y_{0}}}\pm {\frac {r}{d_{0}^{2}}}{\sqrt {d_{0}^{2}-r^{2}}}{\binom {-y_{0}}{x_{0}}}.}$$

• For d₀ < r no tangents exist.
• For d₀ = r point P₀ lies on the circle and there is just one tangent with equation ${\displaystyle x_{0}x+y_{0}y=r^{2}.}$
• In case of d₀ > r there are 2 tangents with equations ${\displaystyle x_{1}x+y_{1}y=r^{2},\ x_{2}x+y_{2}y=r^{2}.}$

Relation to circle inversion: Equation ${\displaystyle ax_{0}=r^{2}}$ describes the circle inversion of point ${\displaystyle (x_{0},0).}$

Relation to pole and polar: The polar of point ${\displaystyle (x_{0},0)}$ has equation ${\displaystyle xx_{0}=r^{2}.}$

A tangential polygon is a polygon each of whose sides is tangent to a particular circle, called its incircle. Every triangle is a tangential polygon, as is every regular polygon of any number of sides; in addition, for every number of polygon sides there are an infinite number of non-congruent tangential polygons.

A tangential quadrilateral ABCD is a closed figure of four straight sides that are tangent to a given circle C. Equivalently, the circle C is inscribed in the quadrilateral ABCD. By the Pitot theorem, the sums of opposite sides of any such quadrilateral are equal, i.e.,

$${\displaystyle {\overline {AB}}+{\overline {CD}}={\overline {BC}}+{\overline {DA}}.}$$

Tangential quadrilateral

This conclusion follows from the equality of the tangent segments from the four vertices of the quadrilateral. Let the tangent points be denoted as P (on segment AB), Q (on segment BC), R (on segment CD) and S (on segment DA). The symmetric tangent segments about each point of ABCD are equal:

$${\displaystyle {\begin{aligned}{\overline {BP}}={\overline {BQ}}=b,&\quad {\overline {CQ}}={\overline {CR}}=c,\\{\overline {DR}}={\overline {DS}}=d,&\quad {\overline {AS}}={\overline {AP}}=a.\end{aligned}}}$$

But each side of the quadrilateral is composed of two such tangent segments

$${\displaystyle {\begin{aligned}&{\overline {AB}}+{\overline {CD}}=(a+b)+(c+d)\\={}&{\overline {BC}}+{\overline {DA}}=(b+c)+(d+a)\end{aligned}}}$$

proving the theorem.

The converse is also true: a circle can be inscribed into every quadrilateral in which the lengths of opposite sides sum to the same value.[2]

This theorem and its converse have various uses. For example, they show immediately that no rectangle can have an inscribed circle unless it is a square, and that every rhombus has an inscribed circle, whereas a general parallelogram does not.

Finding outer tangent. Two circles' outer tangents.

The red line joining the points (x₃, y₃) and (x₄, y₄) is the outer tangent between the two circles. Given points (x₁, y₁), (x₂, y₂) the points (x₃, y₃), (x₄, y₄) can easily be calculated with help of the angle α:

$${\displaystyle {\begin{aligned}x_{3}&=x_{1}\pm r\sin \alpha \\y_{3}&=y_{1}\pm r\cos \alpha \\x_{4}&=x_{2}\pm R\sin \alpha \\y_{4}&=y_{2}\pm R\cos \alpha \\\end{aligned}}}$$

Here R and r notate the radii of the two circles and the angle α can be computed using basic trigonometry. You have α = γ − β with[4]

$${\displaystyle {\begin{aligned}\gamma &=-{\text{atan2}}\left({y_{2}-y_{1}},{x_{2}-x_{1}}\right)\\[2pt]\beta &=\pm \arcsin \left({\frac {R-r}{\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}}\right)\end{aligned}}}$$

where atan2 the 2-argument arctangent.

Outer tangents to two circles

The distances between the centers of the nearer and farther circles, O₂ and O₁ and the point where the two outer tangents of the two circles intersect (homothetic center), S respectively can be found out using similarity as follows:

$${\displaystyle {\frac {dr}{r_{1}-r_{2}}}}$$

Here, r can be r₁ or r₂ depending upon the need to find distances from the centers of the nearer or farther circle, O₂ and O₁. d is the distance O₁O₂ between the centers of two circles.

Inner tangent. The external tangent lines pass through the internal homothetic center.

An inner tangent is a tangent that intersects the segment joining two circles' centers. Note that the inner tangent will not be defined for cases when the two circles overlap.

The bitangent lines can be constructed either by constructing the homothetic centers, as described at that article, and then constructing the tangent lines through the homothetic center that is tangent to one circle, by one of the methods described above. The resulting line will then be tangent to the other circle as well. Alternatively, the tangent lines and tangent points can be constructed more directly, as detailed below. Note that in degenerate cases these constructions break down; to simplify exposition this is not discussed in this section, but a form of the construction can work in limit cases (e.g., two circles tangent at one point).

Let O₁ and O₂ be the centers of the two circles, C₁ and C₂ and let r₁ and r₂ be their radii, with r₁ > r₂; in other words, circle C₁ is defined as the larger of the two circles. Two different methods may be used to construct the external and internal tangent lines.

External tangents

Construction of the outer tangent

A new circle C₃ of radius r₁ − r₂ is drawn centered on O₁. Using the method above, two lines are drawn from O₂ that are tangent to this new circle. These lines are parallel to the desired tangent lines, because the situation corresponds to shrinking both circles C₁ and C₂ by a constant amount, r₂, which shrinks C₂ to a point. Two radial lines may be drawn from the center O₁ through the tangent points on C₃; these intersect C₁ at the desired tangent points. The desired external tangent lines are the lines perpendicular to these radial lines at those tangent points, which may be constructed as described above.

Internal tangents

Construction of the inner tangent

A new circle C₃ of radius r₁ + r₂ is drawn centered on O₁. Using the method above, two lines are drawn from O₂ that are tangent to this new circle. These lines are parallel to the desired tangent lines, because the situation corresponds to shrinking C₂ to a point while expanding C₁ by a constant amount, r₂. Two radial lines may be drawn from the center O₁ through the tangent points on C₃; these intersect C₁ at the desired tangent points. The desired internal tangent lines are the lines perpendicular to these radial lines at those tangent points, which may be constructed as described above.

Let the circles have centres c₁ = (x₁, y₁) and c₂ = (x₂, y₂) with radius r₁ and r₂ respectively. Expressing a line by the equation ${\displaystyle ax+by+c=0,}$ with the normalization ${\displaystyle a^{2}+b^{2}=1,}$ then a bitangent line satisfies:

$${\displaystyle {\begin{aligned}ax_{1}+by_{1}+c&=r_{1}\\ax_{2}+by_{2}+c&=r_{2}\\\end{aligned}}}$$

Solving for (a, b, c) by subtracting the first from the second yields

$${\displaystyle a\Delta x+b\Delta y=\Delta r,\qquad (1)}$$

where

$${\displaystyle \Delta x=x_{2}-x_{1},\quad \Delta y=y_{2}-y_{1}}$$

and

$${\displaystyle \Delta r=r_{2}-r_{1}}$$

for the outer tangent or

$${\displaystyle \Delta r=r_{2}+r_{1}}$$

for the inner tangent.

If ${\textstyle d={\sqrt {(\Delta x)^{2}+(\Delta y)^{2}}}}$ is the distance from c₁ to c₂ we can normalize by

$${\displaystyle X={\frac {\Delta x}{d}},\quad Y={\frac {\Delta y}{d}},\quad R={\frac {\Delta r}{d}}}$$

to simplify equation (1), resulting in the following system of equations:

$${\displaystyle {\begin{aligned}aX+bY&=R,\\a^{2}+b^{2}&=1;\end{aligned}}}$$

solve these to get two solutions (k = ±1) for the two external tangent lines:

$${\displaystyle {\begin{aligned}a&=RX-kY{\sqrt {1-R^{2}}}\\b&=RY+kX{\sqrt {1-R^{2}}}\\c&=r_{1}-(ax_{1}+by_{1})\end{aligned}}}$$

Geometrically this corresponds to computing the angle formed by the tangent lines and the line of centers, and then using that to rotate the equation for the line of centers to yield an equation for the tangent line. The angle is computed by computing the trigonometric functions of a right triangle whose vertices are the (external) homothetic center, a center of a circle, and a tangent point; the hypotenuse lies on the tangent line, the radius is opposite the angle, and the adjacent side lies on the line of centers.

(X, Y) is the unit vector pointing from c₁ to c₂, while R is cos θ where θ is the angle between the line of centers and a tangent line. sin θ is then ${\displaystyle \pm {\sqrt {1-R^{2}}}}$ (depending on the sign of θ, equivalently the direction of rotation), and the above equations are rotation of (X, Y) by ±θ using the rotation matrix:

$${\displaystyle {\begin{pmatrix}R&\mp {\sqrt {1-R^{2}}}\\\pm {\sqrt {1-R^{2}}}&R\end{pmatrix}}}$$

• k = 1 is the tangent line to the right of the circles looking from c₁ to c₂.
• k = −1 is the tangent line to the right of the circles looking from c₂ to c₁.

The above assumes each circle has positive radius. If r₁ is positive and r₂ negative then c₁ will lie to the left of each line and c₂ to the right, and the two tangent lines will cross. In this way all four solutions are obtained. Switching signs of both radii switches k = 1 and k = −1.

Finding outer tangent. Circle tangents.

In general the points of tangency t₁ and t₂ for the four lines tangent to two circles with centers v₁ and v₂ and radii r₁ and r₂ are given by solving the simultaneous equations:

$${\displaystyle {\begin{aligned}(t_{2}-v_{2})\cdot (t_{2}-t_{1})&=0\\(t_{1}-v_{1})\cdot (t_{2}-t_{1})&=0\\(t_{1}-v_{1})\cdot (t_{1}-v_{1})&=r_{1}^{2}\\(t_{2}-v_{2})\cdot (t_{2}-v_{2})&=r_{2}^{2}\\\end{aligned}}}$$

These equations express that the tangent line, which is parallel to ${\displaystyle t_{2}-t_{1},}$ is perpendicular to the radii, and that the tangent points lie on their respective circles.

These are four quadratic equations in two two-dimensional vector variables, and in general position will have four pairs of solutions.

Two distinct circles may have between zero and four bitangent lines, depending on configuration; these can be classified in terms of the distance between the centers and the radii. If counted with multiplicity (counting a common tangent twice) there are zero, two, or four bitangent lines. Bitangent lines can also be generalized to circles with negative or zero radius. The degenerate cases and the multiplicities can also be understood in terms of limits of other configurations – e.g., a limit of two circles that almost touch, and moving one so that they touch, or a circle with small radius shrinking to a circle of zero radius.

• If the circles are outside each other (${\displaystyle d>r_{1}+r_{2}}$), which is general position, there are four bitangents.
• If they touch externally at one point (${\displaystyle d=r_{1}+r_{2}}$) – have one point of external tangency – then they have two external bitangents and one internal bitangent, namely the common tangent line. This common tangent line has multiplicity two, as it separates the circles (one on the left, one on the right) for either orientation (direction).
• If the circles intersect in two points (${\displaystyle |r_{1}-r_{2}|<d<r_{1}+r_{2}}$), then they have no internal bitangents and two external bitangents (they cannot be separated, because they intersect, hence no internal bitangents).
• If the circles touch internally at one point (${\displaystyle d=|r_{1}-r_{2}|}$) – have one point of internal tangency – then they have no internal bitangents and one external bitangent, namely the common tangent line, which has multiplicity two, as above.
• If one circle is completely inside the other (${\displaystyle d<|r_{1}-r_{2}|}$) then they have no bitangents, as a tangent line to the outer circle does not intersect the inner circle, or conversely a tangent line to the inner circle is a secant line to the outer circle.

Finally, if the two circles are identical, any tangent to the circle is a common tangent and hence (external) bitangent, so there is a circle's worth of bitangents.

Further, the notion of bitangent lines can be extended to circles with negative radius (the same locus of points, ${\displaystyle x^{2}+y^{2}=(-r)^{2},}$ but considered "inside out"), in which case if the radii have opposite sign (one circle has negative radius and the other has positive radius) the external and internal homothetic centers and external and internal bitangents are switched, while if the radii have the same sign (both positive radii or both negative radii) "external" and "internal" have the same usual sense (switching one sign switches them, so switching both switches them back).

Bitangent lines can also be defined when one or both of the circles has radius zero. In this case the circle with radius zero is a double point, and thus any line passing through it intersects the point with multiplicity two, hence is "tangent". If one circle has radius zero, a bitangent line is simply a line tangent to the circle and passing through the point, and is counted with multiplicity two. If both circles have radius zero, then the bitangent line is the line they define, and is counted with multiplicity four.

Note that in these degenerate cases the external and internal homothetic center do generally still exist (the external center is at infinity if the radii are equal), except if the circles coincide, in which case the external center is not defined, or if both circles have radius zero, in which case the internal center is not defined.

The internal and external tangent lines are useful in solving the belt problem, which is to calculate the length of a belt or rope needed to fit snugly over two pulleys. If the belt is considered to be a mathematical line of negligible thickness, and if both pulleys are assumed to lie in exactly the same plane, the problem devolves to summing the lengths of the relevant tangent line segments with the lengths of circular arcs subtended by the belt. If the belt is wrapped about the wheels so as to cross, the interior tangent line segments are relevant. Conversely, if the belt is wrapped exteriorly around the pulleys, the exterior tangent line segments are relevant; this case is sometimes called the pulley problem.