Questions: Problem 1: Consider two circles of radius 2: one of them centered at the point (1,0), the other one centered at the point (0,1). In this problem, we'll find algebraically where these circles intersect. (a) Sketch a picture of the two circles. Based on your picture, do you have any predictions about where the two circles intersect? (b) For each circle, write an equation that defines it. (c) Suppose (x, y) is a point that satisfies both equations at once. What does that tell you about x and y? (Hint: try adding or subtracting the two equations to/from each other.) (d) Using your answer to part (c), find the point(s) where the two circles intersect. Does your answer match up with what you see in your picture?

Solution Steps

(b) To find the equations of the circles, use the standard form of a circle's equation \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. For the first circle centered at \((1, 0)\) with radius 2, the equation is \((x - 1)^2 + y^2 = 4\). For the second circle centered at \((0, 1)\) with radius 2, the equation is \(x^2 + (y - 1)^2 = 4\).

(c) If \((x, y)\) satisfies both equations, it lies on both circles. To find such points, solve the system of equations. One approach is to subtract one equation from the other to eliminate terms and simplify the system.

(d) Solve the simplified system of equations to find the intersection points of the circles.

The equations of the two circles are given by:

1. For the circle centered at \((1, 0)\) with radius 2: \[ (x - 1)^2 + y^2 = 4 \]
2. For the circle centered at \((0, 1)\) with radius 2: \[ x^2 + (y - 1)^2 = 4 \]

To find the intersection points, we can subtract the second equation from the first: \[ -y^2 + (x - 1)^2 - (y - 1)^2 + x^2 = 0 \] This simplifies to: \[ -x^2 + y^2 + (x - 1)^2 - (y - 1)^2 = 0 \]

From the simplification, we find that: \[ y = x \] This indicates that the intersection points lie along the line \(y = x\).

Substituting \(y = x\) back into the first circle's equation: \[ (x - 1)^2 + x^2 = 4 \] Expanding and simplifying gives: \[ 2x^2 - 2x + 1 = 4 \implies 2x^2 - 2x - 3 = 0 \] Dividing through by 2: \[ x^2 - x - \frac{3}{2} = 0 \] Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\): \[ x = \frac{1 \pm \sqrt{1 + 6}}{2} = \frac{1 \pm \sqrt{7}}{2} \] Thus, the \(x\) coordinates of the intersection points are: \[ x = \frac{1}{2} - \frac{\sqrt{7}}{2} \quad \text{and} \quad x = \frac{1}{2} + \frac{\sqrt{7}}{2} \] Since \(y = x\), the corresponding \(y\) coordinates are the same.

Final Answer