Math · Precalculus · Worked example

Find where a line meets a circle

Find the points where the line y = x + 1 meets the circle x² + y² = 25.

x^{2} + y^{2} = 25

Substitute the line

The line and the circle form a nonlinear system, and its solutions are their intersection points. Replace y with x + 1 in the equation of the circle.

x^{2} + {(x + 1)}^{2} = 25

Solve the quadratic

Expand and collect: 2x² + 2x − 24 = 0. Divide by 2 and factor.

\begin{array}{l} x^{2} + {(x + 1)}^{2} = 25 \\ 2 x^{2} + 2 x - 24 = 0 \\ x^{2} + x - 12 = 0 \\ (x + 4) (x - 3) = 0 \end{array}

Find the y-values

Use the line: x = 3 gives y = 4, and x = −4 gives y = −3. Both points satisfy the circle’s equation.

\begin{array}{l} 3^{2} + 4^{2} = 25 \\ {(- 4)}^{2} + {(- 3)}^{2} = 25 \end{array}

Result

The intersection points are (3, 4) and (−4, −3).

Your turn

Where does the line y = 7 − x meet the circle x² + y² = 25?

Show the answer and explanation

At (3, 4) and (4, 3).

x² + (7 − x)² = 25 gives 2x² − 14x + 24 = 0, or x² − 7x + 12 = 0, which factors as (x − 3)(x − 4) = 0. Then y = 7 − x gives 4 and 3.

\begin{array}{l} x^{2} + {(7 - x)}^{2} = 25 \\ x^{2} - 7 x + 12 = 0 \\ (x - 3) (x - 4) = 0 \end{array}

Keep exploring

In Graph, the line crosses the circle at both points. Change the line to y = x + 8 and it misses the circle: the quadratic then has a negative discriminant.

Return to the concept →

Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

Make it concrete

Try in the workspace

Open the example inputs, change a value and keep a useful result on your board.

See the line and circle in Graph Check the algebra in Math Open worked example on a board General conic in Math Reference

Your existing work stays on this device. Examples open as editable copies.