Circle Tangent to Circle

Let us model the circle $\Gamma$ as $(x-16)^2 + (y-a)^2 = (a+2)^2$ such that it is tangent to line $y=-2$ at the point $(16,-2).$ If $\Gamma$ is externally tangent to the unit circle $x^2 + y^2 = 1$ , then we have the relation:

$x^2 + y^2 = 1;$

$(x^2 -32x + 256) + (y^2 -2ay + a^2) = a^2 + 4a + 4$ ;

or $y = \frac{-32x - 4a + 253}{2a}$ . If we substitute this expression into the unit circle, then we obtain the quadratic equation:

$x^2 + ( \frac{-32x - 4a + 253}{2a})^2 = 1$ ;

or $4a^2x^2 + (1024x^2 - 64(253-4a)x + (253-4a)^2 = 4a^2;$

or $x = \frac{64(253-4a) \pm \sqrt{64^2(253-4a)^2 - 4(4a^2+1024)[(253-4a)^2 - 4a^2]}}{4a^2 + 1024}$ . We desire only one external tangency point , which occurs the discriminant equals zero:

$64^2(253-4a)^2 - 4(4a^2+1024)[(253-4a)^2 - 4a^2] = 0$ ;

or $(253-4a)^{2}(64^2 - 16a^2 -64^2) + 16a^2 = 0;$

or $16a^2[1 - (253-4a)^2] = 0;$

or $a = 0,\frac{247}{6}, \frac{255}{2}$ . We now check for tangency feasibility between $\Gamma$ and the unit circle. For $a=0$ , we have $(x-16)^2 + y^2 = 4 \Rightarrow$ no intersection points. For $a = \frac{255}{2},$ we have $(x-16)^2 + (y - \frac{255}{2})^2 = (\frac{259}{2})^2 \Rightarrow$ two intersection points. By default, $\boxed{a = \frac{247}{6}}$ is our desired answer.