Step 1 :We are given the circle equation \((x-4)^{2}+y^{2}=1\) and the ellipse equation with vertices at \((4,0)\) and \((10,0)\) and one focus at \((4+2 \sqrt{2}, 0)\). We need to find the points of intersection of the circle and the ellipse.
Step 2 :To find the points of intersection, we need to solve the system of equations formed by the circle and the ellipse. This involves setting the \(y^{2}\) terms equal to each other and solving for \(x\), then substituting the \(x\) values back into one of the original equations to solve for \(y\).
Step 3 :The equation of the circle is \((x-4)^{2}+y^{2}=1\) and the equation of the ellipse is \((x-7)^{2}/9 + y^{2}/2 = 1\).
Step 4 :Setting the \(y^{2}\) terms equal to each other and solving for \(x\) gives us \(x = \frac{65}{17} - \frac{9\sqrt{2}}{17}\) and \(x = \frac{9\sqrt{2}}{17} + \frac{65}{17}\).
Step 5 :Substituting these \(x\) values back into the circle equation to solve for \(y\) gives us \(y = -\frac{\sqrt{118 - 54\sqrt{2}}}{17}\), \(y = \frac{\sqrt{118 - 54\sqrt{2}}}{17}\), \(y = -\frac{\sqrt{54\sqrt{2} + 118}}{17}\), and \(y = \frac{\sqrt{54\sqrt{2} + 118}}{17}\).
Step 6 :Combining the \(x\) and \(y\) solutions into coordinate pairs gives us the points of intersection of the circle and the ellipse.
Step 7 :\(\boxed{\text{Final Answer:}}\) The points of intersection of the circle and the ellipse are \(\left(\frac{65}{17} - \frac{9\sqrt{2}}{17}, -\frac{\sqrt{118 - 54\sqrt{2}}}{17}\right)\), \(\left(\frac{65}{17} - \frac{9\sqrt{2}}{17}, \frac{\sqrt{118 - 54\sqrt{2}}}{17}\right)\), \(\left(\frac{9\sqrt{2}}{17} + \frac{65}{17}, -\frac{\sqrt{54\sqrt{2} + 118}}{17}\right)\), \(\left(\frac{9\sqrt{2}}{17} + \frac{65}{17}, \frac{\sqrt{54\sqrt{2} + 118}}{17}\right)\).