Direct common tangents to the given circles

Question

Find the equation of common tangents to the circles $x^2+y^2-12x-8y+36=0$ and $x^2+y^2-4x-2y+4=0$ touching the circles in the distinct points.

The center of first circle $C_1$ is $(6,4)$. Its radius $r_1=4$.

The center of second circle $C_2$ is $(2,1)$. Its radius $r_2=1$.

Distance between $C_1$ and $C_2$ i.e $C_1C_2=5=r_1+r_2$. That means, circles touch each other externally.

Let the point of intersection of direct common tangents be $P$. And $P$ divides $C_1C_2$ externally in the ratio $4:1$. So $P$ is $(\frac23,0)$.

Now, equation of pair of tangents from an external point is $SS_1=T^2$. Using this on the first circle, I get equations as $y=0$ and $y=8$. Using the formula for second circle, I get equations as $y=0$ and $y=2$.

Now, I have two questions here. a) Why are equations coming out to be different? Since it's the same pair of tangents, shouldn't it be the same? b) The pair of tangents should pass through $P$. But $P$ has y-coordinate as zero. So, why am I getting $y=2$ and $y=8?$

Also, if we solve it differently i.e. by taking the slope of tangents as $m$, and then equating the perpendicular distance from center of second circle, onto the equation of line through P, with radius of that circle, we get $y=0$ and $24x-7y=16$. And if I do this on first circle, I get different equations.

What am I doing wrong here? Also, why am I getting so many equations here?

Answer 1

1. x-axes that is $y=0$ is the one of the tangents passing through two distinct points since $C_1(2;1)$ and $r_1=1$ $C_2(6;4)$ and $r_2=4$

2. Now the equation of the line through both centres is $3x-4y-2=0$ that is the bisector of tangent lines through distinct points. They all intersect at $(\frac{2}{3};0)$

and the other one passing through the point $(\frac{2}{3};0)$ with slope is $m=\frac{24}{7}$ is$$24x-7y-16=0$$ (Slope of line through both centers is $m=tan\alpha=\frac{3}{4}$ and slope of other common tangent line is $m=tan2\alpha=\frac{24}{7}$)

3. Circles are tangent to each other, then equation common tangent through common point of tangency is

$$x^2+y^2-12x-8y+36=x^2+y^2-4x-2y+4$$ $$4x+3y-16=0$$

Answer 2

Consider the line passing through the centers of the circles $$y=\frac{3}{4}x-\frac{1}{2}$$ $x-$axis is tangent to both circles because abs value of the $y-$coordinate of the centers is equal to the radii. Thus the intersection of the $x-$axis and the line of the centers give the point $P$ from where the common tangents are drawn. To solve the system we substitute $y=0$ in the equation of the centers line $y=\frac{3}{4}x-\frac{1}{2}$.

We get $P(2/3,0)$. Now write the equation of the line passing through $P$ with unknown slope $m$. $$y=m\left(x - \frac{2}{3}\right)\to 3 m x-3 y-2m=0$$ The distance of this line from the center $(2,1)$ of one of the circle must be equal to its radius $r=1$. $$\frac{\left|6m-3-2m\right|}{\sqrt{9m^2+9}}=1$$ $$|4m-3|=\sqrt{9m^2+9}\to m (7 m-24)=0\to m_1=0;\;m_2=\frac{24}{7}$$ the common tangents are $$y=0;\;y=\frac{24}{7}x-\frac{16}{7}$$

Answer 3

You have already established there is a point of tangency between the two circles. Here is one way to locate the external tangents. Ask what kind of transformation would map the larger circle to the smaller. You know that the larger circle has radius $r_1 = 4$ and center $C_1 = (6,4)$, while the smaller has radius $r_2 = 1$ and center $C_2 = (2,1)$. So if we apply the transformation $(u,v) = (x/4, y/4)$, what happens to the first circle? The radius becomes $1$ but the center is now $(u,v) = (6/4, 4/4) = (3/2, 1)$. That's close, but not quite. We need to add $1/2$ to the horizontal coordinate. So a contraction of $1/4$ followed by a horizontal translation of $1/2$, or $$(u,v) = \left(\frac{x}{4} + \frac{1}{2}, \frac{y}{4} \right)$$ will map $C_1$ to $C_2$ and $r_1$ to $r_2$. Why does this help us? Because the intersection of the external tangent lines is the unique fixed point of this transformation. That is to say, where the tangent lines intersect, this point does not move when transformed in this way. So if we solve $$x = \frac{x}{4} + \frac{1}{2}, \\ y = \frac{y}{4},$$ we find this fixed point is $$(x,y) = \left(\frac{2}{3}, 0\right).$$ So both lines must pass through this point, hence have equation $$y - 0 = m(x - 2/3),$$ for some slope $m$. This allows us to substitute $x$ for $y$ in either equation of the circle; e.g., for $C_2$, we obtain $$\begin{align} 0 &= x^2 + m^2(x-2/3)^2 - 4x - 2m(x-2/3) + 4 \\ &= (1+m^2)x^2 - \left(\frac{4}{3}m^2 + 2m + 4\right)x + \left(\frac{4}{9}m^2 + \frac{4}{3} m + 4\right). \end{align}$$ We require the discriminant of this quadratic in $x$ to be zero, because otherwise there will not be a single unique intersection point for the tangent line and the circle. To avoid dealing with fractions we multiply the coefficients by $9$: $$a = 9(1+m^2) \\ b = 12m^2 + 18m + 36 \\ c = 4m^2 + 12m + 36.$$ Then we must have $$0 = b^2 - 4ac = (12m^2 + 18m + 36)^2 - 36(m^2 + 1)(4m^2 + 12m + 36) = 36m(24 - 7m).$$ This gives us $m = 0$ and $m = \frac{24}{7}$ as solutions, and our desired tangent lines are $$y = 0, \quad y = \frac{27}{4}\left(x - \frac{3}{2}\right).$$

Answer 4

Over the complex numbers, counting with multiplicity, you get four common tangents. This is because by Bezout the two dual conics, parameterising the tangent lines to your circles, are of degree two, and intersecting them gives the common tangents. All four can be real.

Now, for your specific problem, try assuming the common tangents are of the form $Xx+Yy+1=0:$ Then we can intersect the dual conics $\langle 3X^2+4XY+4X+2Y+1,20X^2+48XY+12X+8Y+1\rangle$ which gives us $\langle (16Y-7)(16Y+3)^2,112X+256Y^2+160Y+49\rangle$ and corresponds to the three lines $-\frac14x-\frac{3}{16}y+1=0$ (doubly) and $-\frac32x+\frac{7}{16}y+1=0.$

The fourth line isn't of the assumed form, we can get it by looking at the line at infinity. That is to assume $Z=0$ in the equations: $\langle 3X^2+4XY+4XZ+2YZ+Z^2,20X^2+48XY+12XZ+8YZ+Z^2,Z\rangle$ which reduces to $\langle 3X^2+4XY,20X^2+48XY\rangle$ from which we get $X=0$ (and $\langle Y,X^2\rangle$ which gives the irrelevant cone point) and $xX+yY+zZ=0$ becomes $y=0.$