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Tangent Lines

Lines Tangent to a Circle

Lines Tangent to Conics

Lines Tangent to Standard Conics

Explorations

Let two points, P and Q, be taken on any locally smooth convex curve, and let the point Q move along the curve nearer and nearer to the point P; then the limiting position of the line PQ, as Q moves up to and ultimately coincides with P, is called the tangent line to the curve at point P. The line through P perpendicular to the tangent line is called the normal to the curve at the point P.

Initialize

To initialize Descarta2D, select the input cell bracket and press SHIFT-Enter.

This initialization assumes that the Descarta2D software has been copied into one of the standard directories for AddOns which are on the Mathematica search path, $Path.

<<Descarta2D`

Lines Tangent to a Circle [Top]

Tangent Through a Point On a Circle

Line tangent to a circle.

Let be a circle and a point on it as shown in Figure [tln:fig15]. We desire to find the equation of the tangent line at . Since the slope of the line joining the center (h,k) and is , the slope of the line tangent to the circle will be the negative reciprocal and the equation of the line tangent to the circle at point becomes (point-slope form)

Since the point is on the circle we also have the equation

Adding Equation ([tln:eq01]) to Equation ([tln:eq02]) results in

or, in a factored form that is easier to remember,

If the circle is centered at the origin, , the equation of the tangent line at is

If the circle is given in general form,

then the tangent line at is

Example: Confirm that the point is on the circle

and find the tangent line at that point.

Solution: The function IsOn2D[point,circle] returns True if the point is on the circle; otherwise, it returns False. TangentLines2D[point,circle] returns a list of lines through the point and tangent to the circle.

IsOn2D[p1=Point2D[{5/2,1+Sqrt[3]/2}],

c1=Circle2D[{2,1},1]]

lns=TangentLines2D[p1,c1] //Simplify

Sketch2D[{p1,c1,lns},PlotRange->{{0,4},{-1,3}}]

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Tangents Through a Point Outside a Circle

Lines through a point, tangent to a circle at the origin.

If the point is outside of the circle there will be two tangent lines from to the circle. Consider the circle and the point D(d,0) in a convenient position as shown in Figure [tln:fig18]. Clearly, the two tangent lines can be determined directly from the normal form of a line as

xcosθ+ysinθ-r=0

where

If the point D is rotated by an angle α about the origin, as shown in Figure [tln:fig22], it will have new coordinates and the tangent lines will also be rotated by α resulting in the two lines

xcos(α+θ)+ysin(α+θ)-r=0

where

Using the standard trigonometric formulas for the sine and cosine of the sum of two angles yields

cos(α+θ) | = | cosαcosθ-sinαsinθ |

. | = | |

. | = |

and

sin(α+θ) | = | cosαsinθ+cosθsinα |

. | = | |

. | = |

Lines through a rotated point, tangent to a circle at the origin.

Lines through a general point, tangent to any circle.

As a final adjustment we translate the geometry so that the center of the circle may be a general location (h,k) as shown in Figure [tln:fig23]. Translating the tangent lines from (0,0) to (h,k) using and yields

xcos(α+θ)+ysin(α+θ)-(r+hcos(α+θ)+ksin(α+θ))=0

where cos(α+θ) and sin(α+θ) are functions of , (h,k), and r and d is the distance between (h,k) and . Notice that after the substitutions are performed no trigonometric functions are present in the formulas.

Tangent Contact Points

Given a circle and a point outside the circle, as shown in Figure [tln:fig23], we desire the coordinates of the points of contact between the circle and the two tangent lines. From the previous section it is clear that when the geometry is in the standard position the coordinates of the contact points are given by

where

If the geometry is rotated and translated to a general position the coordinates of the contact points are given by

where cos(α+θ) and sin(α+θ) have the same formulas as in the previous section.

Example: Find the lines passing through the point (3,-1) and tangent to the circle . Find the coordinates of the points of tangency and plot.

Solution: The function TangentLines2D[point,circle] returns a list of lines through the point and tangent to the circle. TangentPoints2D[point,circle] returns a list of the points of tangency.

p1=Point2D[{3,-1}]; c1=Circle2D[{-1,1},2];

objs={TangentLines2D[p1,c1],TangentPoints2D[p1,c1]}

Sketch2D[{p1,c1,objs}]

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Tangent Line Segment Length

To find the length of the tangent line segment drawn from a given point, , to a circle without computing the point of tangency, the following method can be used. Since in Figure [tln:fig18] is a right triangle the distance D between P and is given by

= | ||

. | = |

Therefore the length of the tangent line segment (squared) is found by substituting the coordinates of the point into the equation of the circle.

Example: Find the length of the tangent line segment from the point (4,3) to the circle .

Solution: Descarta2D does not have a built-in function to compute the length of a tangent line segment. However, a few built-in functions can be combined to apply the technique described in this section. The function Quadratic2D[circle] returns the quadratic equation of a circle. Polynomial2D[quad,{x,y}] substitutes the coordinates (x,y) into the quadratic equation.

c1=Circle2D[{-1,-2},2];

Sqrt[Polynomial2D[Quadratic2D[c1],{4,3}]]

Of course this gives the same result as constructing the tangent points and finding the distance directly.

Distance2D[TangentPoints2D[Point2D[{4,3}],c1][[1]],

Point2D[{4,3}]] //Simplify

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Tangents Parallel to a Line

Lines tangent to a circle.

The equations of the two lines parallel to the line L≡Ax+By+C=0 and tangent to the circle as shown in Figure [tln:fig16] are given by

Notice that the constant coefficient C of the line is not involved in the equations of the tangent lines since only the slope is involved in establishing the parallel condition. The formula is derived by constructing a line that passes through the center (h,k) of the circle with slope m=-A/B. The two tangent lines are then determined by offsetting a distance ±r. Using equations the derivation is

L | ≡ | Ax+By+C=0 |

. | . | . |

≡ | Ax+By-(Ah+Bk)=0 | |

. | . | ax+by-(ah+bk)=0 |

. | . | . |

L' | ≡ | ax+by-(ah+bk±r)=0 |

. | . |

where

are the normalized coefficients of the line.

Tangents Perpendicular to a Line

To find the equations of the two lines, L'', perpendicular to the line

Ax+By+C=0

and tangent to the circle

as shown in Figure [tln:fig16], simply use the line Bx-Ay+C=0 (which is perpendicular to the given line) and apply the formula from the previous section. Once again the value of the constant coefficient C has no bearing on the equations of the resulting lines.

Example: Find the lines tangent to the circle and parallel and perpendicular to the line 2x+3y-1=0 and plot.

Solution: The Descarta2D function TangentLines2D[line,circle,Parallel2D] returns a list of lines parallel to the line and tangent to the circle. The function TangentLines2D[line,circle,Perpendicular2D] returns a list of lines perpendicular to the line and tangent to the circle.

l1=Line2D[2,3,-1]; c1=Circle2D[{3,2},1];

lns={TangentLines2D[l1,c1,Parallel2D],

TangentLines2D[l1,c1,Perpendicular2D]}

Sketch2D[{l1,c1,lns},PlotRange->{{-2,5},{-2,5}}]

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Descarta2D Hint: TangentLines2D[line,circle] returns the same result as

TangentLines2D[line,circle,Parallel2D]

because specifying the keyword Parallel2D is optional.

Example: Using the geometric objects from the previous example, find the points of contact of the four tangent lines.

Solution: The function Point2D[line,circle] will return the point of contact if the line is tangent to the circle.

pts=Map[(Point2D[#,Quadratic2D[c1]])&,

Flatten[lns]] //Simplify

Sketch2D[{l1,c1,lns,pts},

PlotRange->{{-2,5},{-2,5}},

CurveLength2D->20]

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Tangents to Two Circles

Lines tangent to two circles.

Suppose and are two circles and we wish to determine the equations of the lines tangent to both circles. We proceed by finding the equations of tangent lines when the circles are in a special position and then we transform the result to a general position.

Let be a circle, with radius , centered at the origin with equation , and let , with radius , be positioned so that its center is on the +x-axis a distance d from the origin with equation . Since is centered at the origin any line tangent to can be written in the form L≡Ax+By+1=0 because no line tangent to can pass through the origin.

Let and be the distances from the center of and to L, respectively. If L is tangent to the circles then and must equal the radii of the circles, yielding

= | ||

= | ||

= |

where , , and . Simplifying, we have the two equations

= | 1 | |

= |

Solving these two equations for A and B produces four pairs of solutions given by

A | = | |

B | = |

where the sign constants and take on the values ±1 in pairs as shown in the lists. The first pair of solutions gives the external, or direct, tangents and the second pair gives the internal, or transverse, tangents.

We now use the special solution given above to find the tangent lines when the circles are in a general position. Let and be the equations of the two circles. To attain a general positioning we first rotate the lines given in the special solution by an angle θ where and . After the rotation we translate the lines from (0,0) to . Applying these transformations yields the four lines

where

and A and B take the values given as before.

Example: Find the four lines tangent to the circles and . Sketch the external tangents and the internal tangents in separate plots.

Solution: The Descarta2D function TangentLines2D[circle,circle] returns a list of lines tangent to two circles. The first two lines in the list are the external tangents (if returned); the third and fourth lines in the list (if returned) are the internal tangents.

{l1,l2,l3,l4}=

TangentLines2D[c1=Circle2D[{3,0},2],

c2=Circle2D[{-3,0},2]]

{Sketch2D[{c1,c2,l1,l2},PlotRange->{{-6,6},{-3,3}}],

Sketch2D[{c1,c2,l3,l4},PlotRange->{{-6,6},{-3,3}}]}

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Lines Tangent to Conics [Top]

Tangent Through Point on Conic

Suppose we have the general equation of a conic curve given by

The equation of the chord joining any two points, and , on the curve may be written

. | . | |

. | . |

as the equation is clearly first-degree in x and y (the terms above first-degree cancel out), and it is satisfied by the two points and . Putting and , we get the equation of the tangent line

or, expanding,

Adding to both sides will cause the right-hand side to vanish, because satisfies the equation of the curve. Thus the equation of the tangent becomes

= | 0. |

This equation is most easily remembered if we compare it with the equation of the curve and notice that it is derived by replacing and with and , xy with and x and y with and . Whether or not is on the curve, the line represented by Equation ([tln:eq06]) is called the polar of with respect to the curve, and is the pole of the line with respect to the curve.

Example: Find the line tangent to the parabola at the point (4,4).

Solution: The Descarta2D function Line2D[point,conic] returns the polar (line) of a pole (point) with respect to a conic. If the point is on the conic, then the line will be tangent to the conic.

l1=Line2D[p1=Point2D[{4,4}],

crv=Parabola2D[{0,0},1,0]]

Sketch2D[{p1,crv,l1},

CurveLength2D->15,

PlotRange->{{-1,7},{-1,5}}]

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Pole Point and Point of Tangency

Given a line L≡px+qy+r=0 and a conic

we wish to determine the coordinates of the pole point, , of L with respect to Q. The equation of the polar (line) of the pole (point) is derived in general form from Equation ([tln:eq06]) and is given by

If this line and line L are the same line, then the coefficients of the polar line must be equal to some multiple of the coefficients of L yielding the following system of three linear equations in three unknowns

kp | = | |

kq | = | |

kr | = |

Solving theses equations for and the constant k (k is ignored) gives

where

= | ||

= | ||

= |

If the line L is tangent to Q, then is the point of tangency; otherwise, is the pole of the polar L with respect to Q. If is zero, the coordinates of the pole are invalid. This condition occurs when the polar L passes through the center of the conic.

Example: Show that the polar (line) found in the previous example corresponds to the pole (point) of tangency.

Solution: The Descarta2D function Point2D[line,conic] returns the pole (point) of a polar (line) with respect to a conic.

Point2D[l1,crv]

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Line Tangent to a Conic Condition

To find the relationship between the coefficients of a line px+qy+r=0 and a general conic such that the line is tangent to the conic we note that the two intersection points between the line and the conic must be coincident. This condition is equivalent to

. | . | |

. | 2(BD-2AE)qr+2(BE-2CD)pr+2(DE-2BF)pq=0. | . |

The exploration lntancon.html derives this equation.

Example: Find the value of c such that the line 2x+5y+c=0 is tangent to the conic represented by and plot.

Solution: The Descarta2D function TangentEquation2D[line,quad] returns an equation establishing the condition that a line be tangent to the conic represented by the quadratic. Solve this equation for the unknown coefficient c.

Clear[c];

l1=Line2D[2,5,c];

q1=Quadratic2D[2,1,-4,-2,-3,1];

eq1=TangentEquation2D[l1,q1]

ans=Solve[eq1,c]

Sketch2D[{Map[(l1 /. #)&,ans],Loci2D[q1]},

PlotRange->{{-3,3},{-2,2}}]

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Polar of a Conic

As previously shown, if the point is on the conic curve

the equation of the tangent line at is

This equation expresses a relation between the coordinates (x,y) of any point on the tangent line, and those of the point of contact . But the equation, being symmetrical with respect to the coordinates (x,y) and , can be interpreted to represent the line passing through the points of contact from when is not on the curve. Thus the polar, which has the same equation as the tangent line, passes through the points of tangency (when they are real) when the point is not on the curve.

Poles and polars.

Without proof we list the following theorems concerning poles and polars that refer to Figure [tln:fig07].

• If the polar of pole passes through pole , then the polar of passes through .

• If the polars of and intersect at point P, then P is the pole of the line .

• The polar of an exterior point is the line joining the points of contact of the tangents drawn from .

• The polar of an interior point P is the locus of the point of intersection of the tangents at the extremities of every chord through P.

• The polar of a focus is the corresponding directrix.

• There is no (finite) polar of the center of a conic.

Example: Show the inverse functional relationship between the polar and the pole (3,-1) with respect to the quadratic equation .

Solution: The Descarta2D function Line2D[point,quad] returns the polar line of the point with respect to the quadratic. The function Point2D[line,quad] returns the pole (point) of the line with respect to the quadratic.

l1=Line2D[Point2D[{3,-1}],

q1=Quadratic2D[2,3,-1,4,-2,1]]

p1=Point2D[l1,q1]

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Tangents Parallel to a Line

Once again, consider the conic curve whose equation is

and the equation of the tangent line at the point on the conic whose equation is

or, in general form,

To find the lines tangent to Q and parallel to a line the following technique can be used. Let be the desired tangent line (the linear coefficients and of are set equal to the corresponding coefficients of because the lines are parallel). If is to be tangent to Q, then the pole point P of with respect to Q must satisfy the equation for Q. The coordinates of P are functions in the variable , therefore, solving this equation for gives the coefficients of the desired tangent line(s) . The Descarta2D function TangentLines2D[line,quad] implements this technique and can be used to derive the specialized formulas for lines tangent to conics in standard position as presented later in this chapter.

Lines Tangent to Two Conics

The equation relating the coefficients of a quadratic equation

to the coefficients of a line px+qy+r=0 tangent to the quadratic given previously is

2(BD-2AE)qr+2(BE-2CD)pr+2(DE-2BF)pq=0. |

If we select a suitable translation, we can insure that the tangent line does not pass through the origin (i.e. r≠0) and the equation of the tangent line can be written in the form

Now, given two quadratic equations

≡ | ||

≡ |

and using Equation ([tln:eq04]) we can find the coefficients and of the lines tangent to the quadratic by solving two quadratic equations in two unknowns, resulting in equations for four tangent lines. The formulas can be derived in symbolic form, but the results are too unwieldy to be of practical use. Descarta2D, however, can be used to construct such tangents when the problem involves numerical coefficients.

Example: Find the four lines tangent to the ellipses

Plot the ellipses and the tangent lines.

Solution: The Descarta2D function TangentLines2D[curve,curve] returns a list of lines tangent to the two curves.

e1=Ellipse2D[{0,0},4,2,0];

e2=Ellipse2D[{0,0},3,2,Pi/2];

lns=TangentLines2D[e1,e2] //N

Sketch2D[{e1,e2,lns}]

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Line Segments Tangent to Two Conics

Given a line tangent to a conic, the tangent point is the pole point of the line with respect to the conic. Using this relationship the line segments connecting the points of tangency can be determined as illustrated in the next example.

Example: Using the geometric objects from the previous example, find the line segments connecting the contact points of the tangent lines.

Solution: Use the function TangentSegments2D[curve,curve] to construct a list of line segments connecting the contact points of the lines tangent to the two curves.

lnSegs=TangentSegments2D[e1,e2] //N

Sketch2D[{e1,e2,lnSegs}]

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Lines Tangent to Standard Conics [Top]

Lines that are tangent to conics in standard position have particularly simple forms. This section summarizes the equations for these tangent lines for the parabola, ellipse and hyperbola.

Tangents to a Parabola

A line that is parallel to the axis of a parabola intersects the parabola in only one (finite) point; all other lines will cut the parabola in two points real and distinct, real and coincident, or complex. Any line which meets a parabola in two coincident points is a tangent line. Table [tln:tbl01] provides formulas for the line tangent to a parabola in standard form at a point . Table [tln:tbl02] provides the formulas for tangents to a parabola in standard form with a given slope m.

Example: Find the lines through the point (-1,-1) that are tangent to the parabola and plot.

Solution: The function TangentLines2D[point,curve] returns a list of the lines through the point and tangent to the curve. The function Point2D[line,curve] will return the point of tangency for each tangent line.

Clear[x,y];

p1=Point2D[{-1,-1}];

crv1=First[Loci2D[Quadratic2D[(y+1)^2==2(x-1),{x,y}]]];

lns=TangentLines2D[p1,crv1]

Sketch2D[{p1,crv1,lns,Map[Point2D[#,crv1]&,lns]}]

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Tangents to a parabola at a point.

Tangents to a parabola with slope m.

Tangents to an Ellipse

A line that intersects an ellipse in two coincident points is a tangent line. As in the case of the circle, but unlike that of the parabola, there will be two tangents with a given slope. Table [tln:tbl03] provides formulas for the line tangent to an ellipse in standard form at a point . In the formulas a is the length of the semi-major axis of the ellipse and b is the length of the semi-minor axis. Table [tln:tbl04] provides the formulas for tangents to an ellipse in standard form with a given slope m.

Example: Find the lines tangent to the ellipse

rotated counter-clockwise about its center point and passing through the point (4,-1) and plot.

Solution: The function TangentLines2D[point,curve] returns a list of lines through the point tangent to the curve. Point2D[line,curve] will return the tangent point of the line with respect to the curve.

p1=Point2D[{4,-1}];

e1=Ellipse2D[{1,0},3,1,45 Degree] //N;

lns=TangentLines2D[p1,e1] //N

Sketch2D[{p1,e1,lns,Map[Point2D[#,e1]&,lns]}]

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Tangents to an ellipse at a point.

Tangents to an ellipse with slope m.

Tangents to a Hyperbola

A line that intersects a hyperbola in two coincident points is a tangent line. For the hyperbola there will be two tangent lines (real and distinct, coincident with an asymptote, or complex) with a given slope. Table [tln:tbl05] provides formulas for the line tangent to a hyperbola in standard form at a point . In the formulas a is the length of the semi-transverse axis of the hyperbola and b is the length of the semi-conjugate axis. Table [tln:tbl06] provides the formulas for tangents to a hyperbola in standard form with a given slope m. Note that for real tangents with slope m the quantity under the radical must be positive. If, for a given slope, the tangents are real for a particular hyperbola, then the tangents are complex for the conjugate hyperbola.

Example: Find the lines tangent to the hyperbola

rotated counter-clockwise about its center point passing through the point (0,0) and plot.

Solution: The function TangentLines2D[point,curve] returns a list of lines through the point tangent to the curve. Point2D[line,curve] will return the tangent point of the line with respect to the curve.

p1=Point2D[{0,0}];

h1=Hyperbola2D[{1,-1},3,2,30 Degree] //N;

lns=TangentLines2D[p1,h1] //N

Sketch2D[{p1,h1,lns,Map[Point2D[#,h1]&,lns]},

CurveLength2D->15]

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Tangents to a hyperbola at a point.

Tangents to a hyperbola with slope m.

Parallel and Perpendicular Tangents

Using the equations from the tables in the previous sections in the columns labeled TANGENT WITH SLOPE m, we can easily construct lines parallel or perpendicular to a given line and tangent to a given second-degree curve.

Example: Find the lines parallel and perpendicular to the line x+2y-2=0 and tangent to the ellipse

and plot.

Solution: The Descarta2D function TangentLines2D[line,curve,Parallel2D] constructs a list of lines parallel to the given line and tangent to the curve; the function TangentLines2D[line,curve,Perpendicular2D] returns a list of lines perpendicular to the given line and tangent to the curve.

l1=Line2D[1,2,-2];

e1=Ellipse2D[{-1,2},4,2,Pi/2];

lns={TangentLines2D[l1,e1,Parallel2D],

TangentLines2D[l1,e1,Perpendicular2D]}

Sketch2D[{l1,e1,lns},PlotRange->{{-8,6},{-5,7}}]

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Explorations [Top]

Line Tangent to a Circle. [lntancir.html]

Show that the line is tangent to the circle for all values of m.

Line Normal to a Quadratic. [lnquad.html]

Show that the normal line passing through the point on the quadratic whose equation is is given by

where

Eyeball Theorem. [eyeball.html]

The tangents to each of two circles from the center of the other are drawn as shown in the figure. Prove that the chords illustrated are equal in length.

Perpendicular Tangents to a Parabola. [pbtnlns.html]

Show that if and are two lines tangent to a parabola that intersect on the directrix of the parabola, then and are perpendicular to each other.

Tangent to a Parabola with a Given Slope. [pbslp.html]

Show that the line tangent to the parabola with slope m is y=mx+p/m.

Tangent to an Ellipse with Given Slope. [ellslp.html]

Show that the lines tangent to the ellipse with slope m are given by .

Tangent to a Hyperbola with Given Slope. [hypslp.html]

Show that the lines tangent to the hyperbola with slope m are given by .

Tangency Point on Circle. [tancirpt.html]

Show that if a line Ax+By+C=0 is tangent to a circle then the coordinates of the point of tangency are

Monge's Theorem. [monge.html]

Given three circles and the external tangent lines of the circles taken in pairs, show that the three intersection points of the three tangent pairs lie on a straight line.

Line Tangent to a Conic. [lntancon.html]

Show that the relationship between the coefficients of a line px+qy+r=0 tangent to the conic is given by

. | . | |

. | 2(BD-2AE)qr+2(BE-2CD)pr+2(DE-2BF)pq=0. | . |

Normals and Minimum Distance. [normal.html]

Given a point and a quadratic Q, find point(s) P(x,y) on Q such that the line is perpendicular to Q at P. Such a line is called a normal to the quadratic. Use the points P to find the minimum distance from to Q. Assume that and Q are defined numerically.

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