Algebraic Methods for Solving Systems of Non-Linear Equations

Possible Types of Solutions for the Points of Intersection of a Circle and an Ellipse

Figure 6 illustrates possible solution sets for a system of equations involving a circle and an ellipse.

• No solution. The circle and ellipse do not intersect. One shape is inside the other or the circle and the ellipse are a distance away from the other.
• One solution. The circle and ellipse are tangent to each other, and intersect at exactly one point.
• Two solutions. The circle and the ellipse intersect at two points.
• Three solutions. The circle and the ellipse intersect at three points.
• Four solutions. The circle and the ellipse intersect at four points.

Example 3

Solving a System of Nonlinear Equations Representing a Circle and an Ellipse

Solve the system of nonlinear equations.

\(x^2+y^2=26\) (1)

\(3x^2+25y^2=100\) (2)

Solution
Let's begin by multiplying equation (1) by −3, and adding it to equation (2).

\((−3)(x^2+y^2)=(−3)(26)\)

\(−3x^2−3y^2=−78 \)

\(3x2+25y^2=100 \)

\(22y^2=22\)

After we add the two equations together, we solve for \(y\).

\(y^2=1\)

\(y=±\sqrt{1}=±1\)

Substitute \(y=±1\) into one of the equations and solve for \(x\).

\(x^2+(1)^2=26\)

\(x^2+1=26 \)

\(x^2=25 \)

\(x=±\sqrt{25}=±5\)

\(x^2+(−1)^2=26\)

\( x^2+1=26\)

\( x^2=25=±5\)

There are four solutions: \((5,1),(−5,1),(5,−1)\), and \((−5,−1)\). See Figure 7.

Figure 7

Try It #3

Find the solution set for the given system of nonlinear equations.
4x2+y2x2+y2==1310