Common tangents: five useful problem patterns

Turn tangency into an equation

A common tangent touches two curves, possibly at different points. Five useful patterns are: a parabola paired with another graph, two higher-degree graphs, a line touching a quartic twice, two circles, and a parameter chosen to give a specified number of tangents. These are a practical collection of methods, not an exhaustive classification of all tangent problems.

For a differentiable graph $y=f(x)$, the tangent at $x=t$ is

$$y=f'(t)x+f(t)-tf'(t).$$

Match both slope and intercept when two such expressions must describe the same line. For a parabola, substituting a proposed line gives a quadratic, and tangency is its repeated-root condition. For a circle, the distance from its center to the line must equal its radius.

A genuine cubic cannot have one line tangent at two distinct points: the difference between the cubic and line would need two distinct double roots, requiring degree at least four. This observation helps when counting tangents from their contact parameters.

1. A parabola and a cubic

How many lines are tangent to both $y=-2x^2$ and $y=x^3-7x+4$?

At a cubic contact parameter $t$, the tangent is

$$y=(3t^2-7)x-2t^3+4.$$

Intersection with the parabola gives

$$2x^2+(3t^2-7)x-2t^3+4=0.$$

Its discriminant must vanish:

$$\begin{aligned} 0&=(3t^2-7)^2-8(-2t^3+4)\\ &=9t^4+16t^3-42t^2+17\\ &=(t-1)^2(9t^2+34t+17). \end{aligned}$$

The distinct real parameters are

$$t=1,\qquad t=\frac{-17\pm2\sqrt{34}}9.$$

They give distinct cubic tangents by the degree argument above. Hence there are three common tangents. The repeated algebraic root $t=1$ still represents only one line.

If you need their equations, reduce higher powers of the quadratic roots before substituting. For example,

$$-2t^3+4=\left(-\frac29t+\frac{68}{81}\right)(9t^2+34t+17) -\frac{2006}{81}t-\frac{832}{81}.$$

At a root of the quadratic, only the final linear remainder remains.

2. Two cubic graphs

Find the common tangents of $f(x)=x^3-x$ and $g(x)=x^3-x-4$.

With contact parameters $s,t$, their tangent equations are

$$y=(3s^2-1)x-2s^3, \qquad y=(3t^2-1)x-2t^3-4.$$

Matching coefficients gives $s^2=t^2$ and $s^3=t^3+2$. The case $s=t$ is impossible. With $s=-t$, we obtain $t=-1$, $s=1$, and therefore

$$\boxed{y=2x-2.}$$

Matching slopes alone would not be sufficient: parallel tangents need not be the same line.

3. A quartic touched twice

Find a line tangent at two distinct points to

$$y=x^4-2x^2-2x+1.$$

The tangent cannot be vertical, since this polynomial graph has a finite derivative everywhere. Write it as $y=mx+n$ and let its distinct contact coordinates be $\alpha<\beta$. The difference polynomial is monic and has double roots at both points, so

$$x^4-2x^2-(2+m)x+1-n=(x-\alpha)^2(x-\beta)^2.$$

Coefficient comparison gives

$$\alpha+\beta=0,\qquad (\alpha+\beta)^2+2\alpha\beta=-2,$$ $$2\alpha\beta(\alpha+\beta)=2+m,\qquad \alpha^2\beta^2=1-n.$$

Hence $\alpha\beta=-1$, $m=-2$, and $n=0$. The contact coordinates are the roots of $u^2-1=0$, namely $-1$ and $1$, confirming they really are distinct and real. Thus

$$\boxed{y=-2x.}$$

The factorization also checks the result directly: the difference is $(x^2-1)^2$. The area between the curve and this line is

$$\int_{-1}^{1}(x^2-1)^2\,dx=\frac{16}{15}.$$

More generally,

$$\int_\alpha^\beta(x-\alpha)^2(x-\beta)^2\,dx =\frac{(\beta-\alpha)^5}{30}.$$

For a quartic with leading coefficient $a$, the ordinary enclosed area includes the factor $|a|$.

4. Two circles

For a line $ax+by+c=0$, where $(a,b)\ne(0,0)$, the distance from $(x_0,y_0)$ is

$$\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}.$$

Equating this to each radius handles circle tangency and includes vertical lines. Two distinct circles can have up to four common tangents. External tangents place both centers on the same side; internal tangents place them on opposite sides. Coincident circles are an exception with infinitely many tangents.

Example and a correction to the source

The source statement uses $x^2+y^2=1$ and $(x-2)^2+y^2=1$, but its solution uses center $(3,0)$. These are different configurations.

For the stated pair with centers two units apart, the circles touch externally. The complete answer is

$$\boxed{y=1,\qquad y=-1,\qquad x=1.}$$

In particular, excluding vertical lines would lose a valid tangent. To check completeness, a nonvertical line $y=mx+n$ must satisfy $n^2=m^2+1$ and $(2m+n)^2=m^2+1$. Thus $m(m+n)=0$; $m=-n$ contradicts the first equation, leaving $m=0$, $n=\pm1$. A vertical common tangent must be $x=1$.

For the pair actually solved in the source, with second center $(3,0)$, there is no common vertical tangent. The conditions are instead

$$n^2=m^2+1,\qquad(3m+n)^2=m^2+1.$$

Subtracting gives $3m(3m+2n)=0$. Either $m=0$, $n=\pm1$, or $n=-3m/2$, which gives $m=\pm2/\sqrt5$. The four lines are

$$\boxed{y=\pm1,\qquad y=\pm\frac2{\sqrt5}x\mp\frac3{\sqrt5},}$$

with the paired signs taken together.

5. Choose a parameter to obtain four tangents

Find all $c$ for which $y=x^3$ and $y=x^2+x+c$ have four common tangents.

The tangent to the cubic at $t$ is $y=3t^2x-2t^3$. Substituting it into the parabola gives

$$x^2+(1-3t^2)x+c+2t^3=0.$$

The discriminant condition is

$$9t^4-8t^3-6t^2+1=4c.$$

Different real $t$ give different cubic tangents, so we need four distinct real solutions. Let $F(t)=9t^4-8t^3-6t^2+1$. Then

$$F'(t)=12t(t-1)(3t+1).$$

Its successive critical values are

$$F(-1/3)=\frac{20}{27},\qquad F(0)=1,\qquad F(1)=-4.$$

The derivative signs are $-,+,-,+$ on the four intervals, and $F(t)\to+\infty$ at both ends.

A horizontal line meets all four monotone branches exactly when $20/27<4c<1$. At either boundary, two intersections merge and there are fewer than four distinct tangents. Consequently,

$$\boxed{\frac5{27}<c<\frac14.}$$

This is an application of root placement and parameter separation. In general, always justify the relationship between contact parameters and distinct lines before using a root count as a tangent count.