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
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asked Nov 18, 2017 at 14:29
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$\begingroup$ math-only-math.com/circle-touches-x-axis.html $\endgroup$– lab bhattacharjeeNov 18, 2017 at 14:38
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$\begingroup$ My first attempt was same but couldn't continue. Can you elaborate your method. $\endgroup$– Rohan ShindeNov 18, 2017 at 14:43
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1$\begingroup$ 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 $u^2=1/3$. $\endgroup$– LozengesNov 18, 2017 at 16:02
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$\begingroup$ 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, $m = \tan (2\theta)$. $\endgroup$– MickNov 18, 2017 at 16:29
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1$\begingroup$ @Mick $P$ is not on the line of centers. It is a point of intersection of the two circles. $\endgroup$– LozengesNov 18, 2017 at 16:46
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2 Answers
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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}
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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 $(2), m=\dfrac{h^2-a^2}{2ha}$
answered Nov 18, 2017 at 16:29
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$\begingroup$ $\frac{-m h^2}{m}=-h^2=\frac{52}{3}$. something is wrong $\endgroup$– LozengesNov 18, 2017 at 17:27
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$\begingroup$ @Lozenges, I thought so. But I'm yet to pinpoint the fault $\endgroup$ Nov 18, 2017 at 18:25