It's All Circles.

Let $r = \dfrac{85}{6}$ .

Using the points $P$ , $Q$ , and $R$ on the circle above $\implies$

$a^2 + b^2 = r^2$

$(j + 7 - a)^2 + (j - b)^2 = r^2$

$a^2 + (j + 4 + b)^2 = r^2$

Solving the system above we obtain:

$\implies (j + 4)^2 + 2(j + 4)b = 0 \implies \boxed{b = -\dfrac{j + 4}{2}} \implies 3j^2 + 18j + 49 = 2(j + 7)a \implies \boxed{a = \dfrac{3j^2 + 18j + 49}{2(j + 7)}}$

$\implies (\dfrac{3j^2 + 18j + 49}{2(j + 7)})^2 + (\dfrac{j + 4}{2})^2 = (\dfrac{85}{6})^2 \implies\dfrac{10j^4 + 130j^3 + 795j^2 + 2380j + 3185}{4(j^2 + 14j + 49)} = \dfrac{7225}{36} \implies$

$360j^4 + 4680j^3 - 280j^2 - 318920j - 1301440 = 0$

By inspection one solution is $j = 8$ .

Dividing $360j^4 + 4680j^3 - 280j^2 - 318920j - 1301440$ by $j - 8$ we obtain:

$40(j - 8)(9j^3 + 189j^2 + 1505j + 4067) = 0$ and $9j^3 + 189j^2 + 1505j + 4067 > 0$ for
$j > 0 \implies j = \boxed{8}$ .