Problem

Homework
Question 18, 11.3.90
Hiw Score: $66.6 \% \%, 12$ or 18 points
Points: 0 of 1
Consider the circle $(x-4)^{2}+y^{2}=1$ and the ellipse with vertices at $(4,0)$ and $(10,0)$ and one focus at $(4+2 \sqrt{2}, 0)$. Find the points of intersection of the circle and the ellipse.

The points of intersection of the circle and the ellipse are $\square$.
(Type ordered pairs. Use integers or fractions for any numbers in the expression. Type an exact answer, using radicals as needed. Use a comma to separate answers as needed.)

Answer

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Answer

\(\boxed{\text{Final Answer:}}\) The points of intersection of the circle and the ellipse are \(\left(\frac{65}{17} - \frac{9\sqrt{2}}{17}, -\frac{\sqrt{118 - 54\sqrt{2}}}{17}\right)\), \(\left(\frac{65}{17} - \frac{9\sqrt{2}}{17}, \frac{\sqrt{118 - 54\sqrt{2}}}{17}\right)\), \(\left(\frac{9\sqrt{2}}{17} + \frac{65}{17}, -\frac{\sqrt{54\sqrt{2} + 118}}{17}\right)\), \(\left(\frac{9\sqrt{2}}{17} + \frac{65}{17}, \frac{\sqrt{54\sqrt{2} + 118}}{17}\right)\).

Steps

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 \(\left(\frac{65}{17} - \frac{9\sqrt{2}}{17}, -\frac{\sqrt{118 - 54\sqrt{2}}}{17}\right)\), \(\left(\frac{65}{17} - \frac{9\sqrt{2}}{17}, \frac{\sqrt{118 - 54\sqrt{2}}}{17}\right)\), \(\left(\frac{9\sqrt{2}}{17} + \frac{65}{17}, -\frac{\sqrt{54\sqrt{2} + 118}}{17}\right)\), \(\left(\frac{9\sqrt{2}}{17} + \frac{65}{17}, \frac{\sqrt{54\sqrt{2} + 118}}{17}\right)\).

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