Examples and Solutions
Example 1
Find the points of intersection of the circles given by their equations as follows:
\[ (x - 2)^2 + (y - 3)^2 = 9 \] \[ (x - 1)^2 + (y + 1)^2 = 16 \]View Solution
The points of intersections are found by solving the above system of equations.
We first expand the two equations as follows:
\[ x^2 - 4x + 4 + y^2 - 6y + 9 = 9 \] \[ x^2 - 2x + 1 + y^2 + 2y + 1 = 16 \]Multiply all terms in the first equation by -1 to obtain an equivalent equation and keep the second equation unchanged:
\[ -x^2 + 4x - 4 - y^2 + 6y - 9 = -9 \] \[ x^2 - 2x + 1 + y^2 + 2y + 1 = 16 \]We now add the same sides of the two equations to obtain a linear equation:
\[ 2x - 3 + 8y - 8 = 7 \]Which may be written as:
\[ x + 4y = 9 \quad \text{or} \quad x = 9 - 4y \]We now substitute \(x\) by \(9 - 4y\) in the first equation to obtain:
\[ (9 - 4y)^2 - 4(9 - 4y) + 4 + y^2 - 6y + 9 = 9 \]Which may be written as:
\[ 17y^2 - 62y + 49 = 0 \]Solve the above quadratic equation for \(y\) to obtain two solutions:
\[ y = \frac{31 + 8\sqrt{2}}{17} \approx 2.49 \quad \text{and} \quad y = \frac{31 - 8\sqrt{2}}{17} \approx 1.16 \]We now substitute the values of \(y\) already obtained into the equation \(x = 9 - 4y\) to obtain the values for \(x\):
\[ x = \frac{29 + 32\sqrt{2}}{17} \approx -0.96 \quad \text{and} \quad x = \frac{29 - 32\sqrt{2}}{17} \approx 4.37 \]The two points of intersection of the two circles are given by:
\[ (-0.96, 2.49) \quad \text{and} \quad (4.37, 1.16) \]Shown below is the graph of the two circles and the linear equation \(x + 4y = 9\) obtained above:
