Dynamic Geometry: P93

Let the $x$ -axis be the straight line tangent to both cyan and green circles. And the $y$ -axis be in the middle of the center $O$ of the cyan circle and the center $Q$ of the green circle, and an arbitrary point on the locus, the center of the yellow circle, be $P(x,y)$ while its radius is $r$ .

Then by cosine rule ,

$\begin{aligned} PQ^2 & = OP^2+OQ^2 - 2\cdot OP \cdot OQ \cdot \cos \angle POQ \\ (r+1)^2 & = (r+4)^2 + 5^2 - 10\cdot OP \cos \angle POQ \\ \implies OP \cdot \cos \angle POQ & = \frac {(r+4)^2 + 5^2 - (r+1)^2}{10} = \frac {3r+20}5 \end{aligned}$

Note that $OP \cdot \cos \angle POQ = ON$ is the projection of $OP$ on $OQ$ . Also note that the gradient of $ON$ is $- \tan \frac 34$ and the coordinates of $N(x_N, y_N)$ are $x_N = \dfrac 45 \cdot ON - 2 = \dfrac {12r+30}{25}$ and $y_N = 4 - \dfrac 35 \cdot ON = \dfrac {40-9r}{25}$ .

By Pythagorean theorem ,

$\begin{aligned} PN^2 & = OP^2 - ON^2 = (r+4)^2 - \left(\frac {3r+20}5 \right)^2 = \frac {16r^2+80r}{25} \\ \implies PN & = \frac {4\sqrt{r^2+5r}}5 = 2 \blue{\tan \phi} \\ \implies \tan \phi & = \frac {2\sqrt{r^2+5r}}5 \\ \sec^2 \phi & = \tan^2 \phi + 1 = \frac {4r^2+20r + 25}{25} = \left(\frac {2r+5}5 \right)^2 \\ \implies \sec \phi & = \frac {2r+5}5 \\ x_N & = \frac {12r+30}{25} = \frac 65 \cdot \frac {2r+5}5 = \frac 65 \sec \phi \\ y_N & = \dfrac {40-9r}{25} = \frac {40}{25} - \frac 9{10} \cdot \frac {2r+5}5 + \frac {45}{50} = \frac 52 - \frac 9{10} \sec \phi \end{aligned}$

The coordinates of $P(x,y)$ :

$\begin{cases} x = x_N + \dfrac 35 \cdot PN = \dfrac 65 \sec \phi + \dfrac 65 \tan \phi \\ y = y_N + \dfrac 45 \cdot PN = \dfrac 52 - \dfrac 9{10} \sec \phi + \dfrac 85 \tan \phi \end{cases} \\ \begin{cases} 3x + 4y & = 10 + 10 \tan \phi & \implies \tan \phi = \dfrac {3x+4y-10}{10} \\ 8x - 6y & = - 15 + 15 \sec \phi & \implies \sec \phi = \dfrac {8x-6y+15}{15} \end{cases} \\ \begin{aligned} 1 + \tan^2 \phi & = \sec^2 \phi \\ 1 + \left(\frac {3x+4y-10}{10}\right)^2 & = \left(\frac {8x-6y-15}{15}\right)^2 \\ 1 + \frac {9x^2+24xy+16y^2-60x-80y+100}{100} & = \frac {64x^2-96xy+36y^2+240x-180y+225}{225} \\ \implies 7x^2-24xy+60x-36 & = 0 \\ \implies y & = \frac 7{24}x + \frac 52 - \frac 3{2x} \end{aligned}$

The locus is part of a hyperbola. Note $y$ of the locus is $r$ of the yellow circle, The smallest radius (red vertical line) is given by $\dfrac 1{\sqrt r_{\min}} = \dfrac 1{\sqrt 1} + \dfrac 1{\sqrt 4} = \frac 32 \implies r_{\min} = \dfrac 49$ . From $\dfrac 7{24}x + \dfrac 52 - \dfrac 3{2x} = \dfrac 49 \implies x_{\min} = \dfrac 23$ . The largest radius (purple vertical line) $r_{\max} = 4$ , when the yellow circle is congruent with the cyan circle; and the $x_{\max} = 6$ . The area under the locus is

$\begin{aligned} A & = \int_\frac 23^6 y \ dx \\ & = \int_\frac 23^6 \frac 7{24}x + \frac 52 - \frac 3{2x} \ dx \\ & = \frac 7{48}x^2 + \frac 52 x - \frac 32 \ln x \ \bigg|_\frac 23^6 \\ & = \frac {500}{27} - \ln 27 \end{aligned}$

Therefore $p+q = 500+27 = \boxed{527}$ .

The center (x,y) and radius $r$ of the moving circle, is governed by these two equations $(x-4)^{2}+(y-4)^{2}=(4+r)^{2}$ and $(x-8)^{2}+(y-1)^{2}=(1+r)^{2}$ .

$y$ solves to $\frac{-7x^{2}+24x+144}{144-24x} = \frac{7x}{24}-\frac{3}{2(x-6)}+\frac{3}{4}$

$\int_ \frac{20}{3}^{12} f(x)dx = \frac{500}{27}-\ln 27$