# Finding slope of line of tangent of two circles

Let $S_1=0$ and $S_2=0$ be the two circles intersecting at point $P(6,4)$ and both tangent to x-axis and line $y=mx$(where $m\gt 0$). If product of radii of the circles $S_1=0$ and $S_2=0$ is $\frac{52}{3}$ then find value of m where m was the slope of line. I tried writing of equations of tangents i.e the X axis and the given line with respect to the radius which will be equal to the y coordinate of the circles but I couldn't use the condition of product in it. I also tried using the parametric form but got messed up. Thanks in advance

• math-only-math.com/circle-touches-x-axis.html Nov 18, 2017 at 14:38
• My first attempt was same but couldn't continue. Can you elaborate your method. Nov 18, 2017 at 14:43
• Hint: Centers $(a,b)$ and $(c,d)$ lie on the line $y=ux$, so $b=ua$ and $d=uc$. Common point $(6,4)$ says $a$ and $c$ are the roots of the same quadratic equation with constant term $=52$. Therefore $u^2=1/3$. Nov 18, 2017 at 16:02
• Let P = (6, 4) and $\theta = \angle POx$. Both O and P are points on the line of centers. Then, $\theta$ is given by $\tan \theta = \dfrac {4}{6}$. Hence, $m = \tan (2\theta)$.
– Mick
Nov 18, 2017 at 16:29
• @Mick $P$ is not on the line of centers. It is a point of intersection of the two circles. Nov 18, 2017 at 16:46

Let $m=\tan2\phi$. Using the general equation for the circle $\mathcal{C}((x_c,y_c),r)$ \begin{align} (x-x_c)^2+(y-y_c)^2-r^2&=0 \end{align}
for the two circles $\mathcal{C}_1((x_1,y_1),r_1)$ and $\mathcal{C}_2((x_2,y_2),r_2)$ with the common point $P=(x,y)=(6,4)$, we have a system \begin{align} (6-r_1\cot\phi)^2+(4-r_1)^2-r_1^2&=0 ,\\ (6-r_2\cot\phi)^2+(4-r_2)^2-r_2^2&=0 ,\\ \end{align}

which gives a solution for $r_1,r_2$ in terms of $\tan\phi$: \begin{align} r_1 &= \tan\phi(6+4\,\tan\phi-4\,\sqrt{\tan^2\phi+3\,\tan\phi-1} ,\\ r_2 &= \tan\phi(6+4\,\tan\phi+4\,\sqrt{\tan^2\phi+3\,\tan\phi-1} . \end{align}

Next, \begin{align} r_1\cdot r_2 &=52\,\tan^2\phi=\tfrac{52}3 ,\\ \phi&=\tfrac\pi6 ,\\ m&=\tan\tfrac\pi3=\sqrt3 . \end{align}

From this, the equation of any circle touching $x$ axis can be written as

$$(x-h)^2+(y-a)^2=a^2\ \ \ \ (1)$$

Now $y=mx$ will be tangent of this iff $$|a|=\dfrac{|mh-a|}{\sqrt{1+m^2}}$$

Squaring we get $$a^2(1+m^2)=(mh-a)^2$$

As $m\ne0,$

$$a^2m+2ha-mh^2=0\ \ \ \ (2)$$

The absolute values of the two roots represent the two radii.

$\implies\dfrac{-mh^2}m=\dfrac{52}3$

Use the fact that $(1)$ passes through $(6,4)$ to find $a$

By $(2), m=\dfrac{h^2-a^2}{2ha}$

• $\frac{-m h^2}{m}=-h^2=\frac{52}{3}$. something is wrong Nov 18, 2017 at 17:27
• @Lozenges, I thought so. But I'm yet to pinpoint the fault Nov 18, 2017 at 18:25