Common tangent to two circles

Find the equations of the common tangents to the 2 circles:

$$(x - 2)^2 + y^2 = 9$$ and
$$(x - 5)^2 + (y - 4)^2 = 4.$$

I've tried to set the equation to be $y = ax+b$, substitute this into the 2 equations and set the discriminant to zero, we then get a simultaneous quadratic equations. But they are really difficult to solve. So is there any simpler way to do this? Thank you.

Solution 1:

A line with equation $ax+y+c=0$ is tangent to a circle iff its distance from the center of the circle is equal to the radius of the circle, so you get the system of equations:

$$\left\{\begin{array}{rcl}|2a+c| &=& 3\sqrt{a^2+1} \\ |5a+4+c| &=& 2\sqrt{a^2+1}. \end{array}\right.$$

with $4$ solutions.

Solution 2:

You can approach this with homogeneous coordinates. A circle with equation $(x-x_c)^2+(y-y_c)^2 - r^2 = 0$ is represented by a 3x3 matrix as

$$ {\rm Circle}(x_c,y_c,r) = \begin{bmatrix} 1 & 0 & -x_c \\ 0 & 1 & -y_c \\ -x_c & -y_c & x_c^2+y_x^2-r^2 \end{bmatrix} $$

This means that the equation for a circle $C_1 = \begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & 0 \\ -2 & 0 & -5 \end{bmatrix}$ is given by the quadratic form

$$ P^\top C_1 P = 0 $$ $$ x^2-4 x+y^2-5 = 0 $$

which is the equation for the first circle when expanded out, and $P=\begin{pmatrix} x&y&1 \end{pmatrix} ^\top $ is an arbitrary point.

The second circle is $ C_2 = \begin{bmatrix} 1 & 0 & -5 \\ 0 & 1 & -4 \\ -5 & -4 & 37 \end{bmatrix} $.

Now here is the fun stuff. A line in this notation in general is defined as $L=\begin{vmatrix}a&b&c\end{vmatrix}^\top$ such that the equation of the line is

$$ P^\top L =0 $$ $$ a x+b y+c = 0 $$

Actually $a$, $b$ above designate the direction of the line such that if the line makes an angle $\theta$ with the horizontal then the line is $L=\begin{vmatrix}-\sin\theta&\cos\theta&-d\end{vmatrix}^\top$ and $d$ is the distance of the line to the origin.

We are using the above information to find the lines that are tangent to both circles. A tangent line to the first circle has satisfies the equation

$$ L^\top C_1^{-1} L =0 $$ $$ d = \pm 3 -2 \sin \theta $$

with the two possible line equations

$$ L_A = \begin{vmatrix}-\sin\theta_A&\cos\theta_A &2\sin\theta_A-3\end{vmatrix}^\top $$ $$ L_B = \begin{vmatrix}-\sin\theta_B&\cos\theta_B &2\sin\theta_B+3\end{vmatrix}^\top $$

to find the orientations of these lines $\theta_A$ and $\theta_B$ we have to find the lines that are tangent to the second circle, and match the coefficients

$$ L^\top C_2^{-1} L =0 $$ $$ d = \pm 2 + 4 \cos \theta -5 \sin \theta $$

with also two possible line equations

$$ L_A = \begin{vmatrix}-\sin\theta_A&\cos\theta_A &-4\cos\theta_A+5\sin\theta_A-2\end{vmatrix}^\top $$ $$ L_B = \begin{vmatrix}-\sin\theta_B&\cos\theta_B &-4\cos\theta_B+5\sin\theta_B+2\end{vmatrix}^\top $$

Setting $L_A=L_A$ and solving for $\theta_A$ yields the following

$$ 2\sin\theta_A-3 = -4\cos\theta_A+5\sin\theta_A-2 $$ $$ 4\cos\theta_A-3\sin\theta_A =1 $$ $$ \sin\theta_A = \frac{8\sqrt{6}-3}{25} $$

with the solution

$$ L_A = \begin{vmatrix} \frac{3-8\sqrt{6}}{25} & \frac{4+6\sqrt{6}}{25} & \frac{16 \sqrt{6}-81}{25} \end{vmatrix} $$ $$ \left( \frac{3-8\sqrt{6}}{25}\right) x + \left(\frac{4+6\sqrt{6}}{25}\right) y + \left(\frac{16 \sqrt{6}-81}{25}\right) = 0 $$ $$ -0.664 x + 0.7479 y - 1.6723 = 0 $$

Similarly with $L_B=L_B$ yielding $4\cos\theta_B-3\sin\theta_B=-1$ or

$$ L_B = \begin{vmatrix} -\frac{3+8\sqrt{6}}{25} & \frac{6\sqrt{6}-4}{25} & \frac{16 \sqrt{6}+81}{25} \end{vmatrix} $$ $$ -\left(\frac{3+8\sqrt{6}}{25}\right) x + \left(\frac{6\sqrt{6}-4}{25}\right) y + \left(\frac{16 \sqrt{6}+81}{25}\right) = 0 $$ $$ -0.904 x + 0.4278 y + 4.808 = 0 $$

Here are the results plotted in GeoGebra for validation.

Solution 3:

The parametric equation of $(x-a)^2+(y-b)^2=r^2$ is $(x=a+r\cos C,y=b+r\sin C)$

Let $$\frac{y-(b+r\sin C)}{x-(a+r\cos C)}=m$$ be a tangent at $(a+r\cos C,b+r\sin C)$, then the distance of line from the center is equal to the radius.

$$r=\frac{\mid m(a-a-r\cos C)-b+b+r\sin C\mid}{\sqrt{m^2+1}}=\frac{r\mid\sin C-m\cos C\mid}{\sqrt{m^2+1}}$$

$m^2+1=(\sin C-m\cos C)^2$

$\implies (m\sin C+\cos C)^2=0\implies m=-\frac{\cos C}{\sin C}$

So, the equation of the tangent becomes

$$\frac{y-(b+r\sin C)}{x-(a+r\cos C)}=-\frac{\cos C}{\sin C}$$

$x\cos C+y\sin C-a\cos C-b\sin C -r=0$ (this can also be reached using calculus)

For $(x-2)^2+y^2=9,a=2,b=0,r=3$

So, equation of the tangent will be $x\cos A+y\sin A-2\cos A -3=0$

For $(x-5)^2+(y-4)^2=4,a=5,b=4,r=2$

So, equation of the tangent will be $x\cos B+y\sin B-5\cos B-4\sin B -2=0$

For common tangent, these two lines must be same,

So, $$\frac{\cos A}{\cos B}=\frac{\sin A}{\sin B}=\frac{2\cos A+3}{5\cos B+4\sin B+2}$$

$\frac{\cos A}{\cos B}=\frac{\sin A}{\sin B}\implies \sin(A-B)=0$

$\implies A=B$ or $A=\pi+B$

(1)If $A=B,1=\frac{2\cos B+3}{5\cos B+4\sin B+2}\implies 4\sin B+3\cos B=1$

(2)If $A=\pi+B,\cos A=\cos(\pi+B)=-\cos B, -1=\frac{-2\cos B+3}{5\cos B+4\sin B+2}$

For (1), $4\sin B+3\cos B=1$

Putting $4=R\sin D,3=R\cos D\implies R=5,D=cos^{-1}\frac 3 5,$

$\cos(B-D)=\frac 1 5, B-\cos^{-1}\frac 3 5 =2n\pi \pm \cos^{-1}\frac 1 5$ where $n$ is any integer.

$\cos B=\frac{3 \pm 8\sqrt{6}}{25},\sin B$ can be calculated uniquely using (1).

So, there will be two tangent in this case.

For(2) $ 3\cos B+4\sin B=-5$

Applying the same approach like in (1), $\cos(B-\cos^{-1}\frac 3 5)=-1=\cos \pi, B=\cos^{-1}\frac 3 5+\pi$ $\cos B=\cos(\cos^{-1}\frac 3 5+\pi)=-\cos(\cos^{-1}\frac 3 5)=-\frac 3 5$

$\sin B$ becomes $-\frac 4 5$

So, the tangent becomes, $(x-5)(-\frac 3 5)+(y-4)(-\frac 4 5) -2=0$

Solution 4:

For the circles $$(x - 2)^2 + y^2 = 9\text{ and } (x - 5)^2 + (y - 4)^2 = 4,$$

using Article $180$ of The elements of coordinate geometry by Loney,

Case $1:$ If $T_2$ is the point dividing internally the line joining the centres in the ratio $2:3$ then its coordinates are $$\left(\frac{2\cdot2+5\cdot3}{2+3},\frac{2\cdot0+4\cdot3}{2+3}\right)=\left(\frac{19}5,\frac{12}5\right)$$

Now, the equation of any line passing through $T_2\left(\frac{19}5,\frac{12}5\right)$ is $$\frac{y-\frac{12}5}{x-\frac{19}5}=m\implies 5mx-5y+12-19m=0$$ where $m$ is the gradient

If it is the tangent of the circle, the distance of this line from the centre of each circle will be equal to the radius of the respective circle.

$$\implies \left|\frac{5m\cdot 2-5\cdot0+12-19m}{\sqrt{(5m)^2+(-5)^2}}\right|=3\implies |12-9m|=15\sqrt{m^2+1}$$

$$\text{On squaring and re-arrangement, } 16m^2+24m+9=0\implies m=-\frac34$$

So, there will be only one crossed common tangent

Case $2:$ If $T_1$ is the point dividing externally the line joining the centres in the ratio $2:3$ then its coordinates are $$\left(\frac{2\cdot2-5\cdot3}{2-3},\frac{2\cdot0-4\cdot3}{2-3}\right)=\left(11,12\right)$$

Now, the equation of any line passing through $T_1\left(11,12 \right)$ is $$\frac{y-12}{x-11}=n\implies nx-y+12-11n=0$$ where $n$ is the gradient

Can you follow the method used in Case $1$ to find the two values of $n,$ which implies there will be two simple common tangents

Solution 5:

Given two circles, determine the internal and external tangents.

Determine a line through the center of the circles. At the center of both circles determine a perpendicular line Note the points where the vertical lines intersect the circle.

TO draw external tangents pick a radius from each circle that is perpendicular to the line through the centers and have both radii on the same side of the center line. Draw a line and determine the equation of the line using where the radii hit the circle. The new line through these two points will hit the line through the center of the circles beyond the smaller circle. The point you just determined is a Vanishing Point. Erase all of the extra lines and points except for the circles, line through the centers and vanishing point. Use the vanishing point to draw the tangents to the nearby circle and continue that line on to the far circle. These same lines will also be tangent to the far circle.

For an internal bisector, draw the perpendicular to the line through the center of each circle This time pick a radius along this perpendicular line where one radius is on each side of the line through the centers. Determine the points where the radii you have drawn hits their respective circles. Draw a line from each of these two points. This line will cross the line through the centers. Mark clearly this point where the line crosses the center line. Use this point to draw the tangent to either circle. The line continued on through the point on the center line to the opposite circle will also be a tangent. Again the problem now boils down to drawing a tangent line to a circle given a point external to the circle that the tangent must pass through.

Make SURE when you use your point on the line through the centers of the circle that you have erased the perpendicular lines AND the points on the circle circumferences where the perpendicular intersected the circle. These points on the circle were NOT the point where the tangent line hits a circle, but only a step used to find the required point on the center line.

Lets say you have marked a point A on the center line and want to use it to find the tangent line from there to a circle centered around point C. It does not matter for this if point A is between the circles or on the far side of the circles. It MUST be the point you marked earlier.

Determine the midpoint M between point A and C.

Write the equation for the circle at center M with radius AM which is equal also to MC. This circle will intersect your circle with center at C at two points. A line from A to one of these points is a tangent line to the circle.

Lets pick one of these points on your circle that you found that will be the point of intersection of your tangent line and call it point P.

Draw lines AP, PC, and MP. Because of your circle with center at M, AM, MP and MC are all equal to one another. This means triangle AMP and triangle PMC are both isosceles with base angles equal to one another. Lets call the base angles in triangle AMP angle 1 and the base angles in triangle PMC angle 2. In triangle APC you find 2 times angle 1 and two times angle 2 in the triangle. The interior angles of a triangle sum to 180. Divide this equation by 2 and you have angle 1 plus angle 2 sum to 90. This makes angle APC a 90 degree angle, so the angle between the line AP and the radius is 90 degrees. This is a Tangent line.

Once you have a tangent line to one of the circles, the same tangent line can be used for the other circle. You now have a circle and a line and need a point of intersection ie solve the line equation for y and plug that into the circle equation and solve for x. This usually involves a quadratic equation with 2 solutions. Use the one on the side of the line through the circle centers that you wish.

Once you have determined the point where the tangent line meets the circle, a perpendicular line to the tangent line though the point in question should go through the center of that circle if you have made no error.

This is a slow method but the steps do not involve math beyond high school analytic geometry. This includes NOT having trigonometry in this method.

For external tangents the idea that objects may be drawn larger or smaller using a vanishing point is more often used and known in art. We found the vanishing point using radii perpendicular to the center line. These lines however were not special for this purpose. Lines at ANY given angle from the center line could have been used to get this same vanishing point. All that matters is that the one of the two objects is a copy of the other albeit magnified.

For internal tangents we used the idea that we may use a focal point between objects to invert an image and or make that image larger or smaller. The images are copies of one another though in this case they are of different sizes. Again we used lines 90 degrees from the center line, but lines at any angle from the center line could have been used. As long as you consistently pick the same angle in both circles, the points on the circumference of each circle should when connected lead toward the same point on the center line.

Review the idea of homeothety in a text book such as Keselev Geometry ie a text for high school students.