Example and Solution
Example
Find the points of intersection of the circle and line:
\[ (x-2)^2 + (y+3)^2 = 4 \] \[ 2x + 2y = -1 \]View Solution
- Solve the linear equation for \( y \): \[ 2x + 2y = -1 \Rightarrow y = -x - \frac{1}{2} \]
- Substitute into the circle equation: \[ (x-2)^2 + \left(-x-\frac{1}{2} +3\right)^2 = 4 \]
- Expand and simplify: \[ 2x^2 - 9x + \frac{25}{4} = 0 \]
- Solve the quadratic equation: \[ x = \frac{9 \pm \sqrt{31}}{4} \]
- Substitute each value of \(x\) into \(y = -x - \frac{1}{2}\): \[ y = \frac{-11 \mp \sqrt{31}}{4} \]
- Therefore, the two points of intersection are: \[ \left(\frac{9+\sqrt{31}}{4},\frac{-11-\sqrt{31}}{4}\right), \qquad \left(\frac{9-\sqrt{31}}{4},\frac{-11+\sqrt{31}}{4}\right) \]
- Approximate values: \[ (3.64,-4.14), \qquad (0.86,-1.36) \]
Shown below is the graph of the circle, the line, and their intersection points.
