find the equation of circle passing through the points of contact of direct common tangents of x^2+y^2=16 and x^2+y^2-12x+32=0

1. The circle \(\displaystyle c_1: x^2+y^2=4^2\) has the center (0, 0) and the radius 4.

The circle \(\displaystyle c_2:x^2+y^2-12x+32=0~\implies~(x-6)^2+y^2=2^2\) has the center (6, 0) and the radius 2.

2. Use similar right triangles to determine the missing lengthes. I used Euklid's theorem to calculate the coordinates of the tangent points: \(\displaystyle T_1\left(\frac43\ ,\ \frac83 \sqrt{2}\right)\)

and \(\displaystyle T_2\left(\frac{20}3\ ,\ \frac43 \sqrt{2}\right)\)

3. The center of the resulting circle \(\displaystyle c_r\) must be on the x-axis and on the perpendicular bisector of \(\displaystyle \overline{T_1T_2}\)

The slope of this perpendicular bisector equals the slope of \(\displaystyle \overline{OT_1} = \dfrac{\frac83 \sqrt{2}}{\frac43}=2\sqrt{2}\)

The midpoint of \(\displaystyle \overline{T_1T_2}\) has the coordinates \(\displaystyle M\left(\dfrac{\frac43 + \frac{20}3}2\ , \ \dfrac{\frac83 \sqrt{2} + \frac{4}3 \sqrt{2}}2 \right)\). That means \(\displaystyle M(4, 2\sqrt{2})\)

4. The perpendicular bisector through M has the equation:

\(\displaystyle y-2 \sqrt{2} = 2\sqrt{2}(x-4)\)

This line crosses the x-axis at:

\(\displaystyle -2 \sqrt{2} = 2\sqrt{2}(x-4)~\implies~-1=x-4~\implies~x=3\)

5. The center of the resulting circle is C(3, 0). Now calculate the distance \(\displaystyle |\overline{CT_1}|\) to get the radius of \(\displaystyle c_r\). I've got \(\displaystyle r=\sqrt{17}\). Therefore the equation of \(\displaystyle c_r\) is:

\(\displaystyle c_r: (x-3)^2+y^2=17\)