Systems of a Line and a Parabola or Circle

Quick answer

A line can cross a parabola twice, touch it once, or miss it entirely, and the same is true of a line and a circle. Substituting the line into the quadratic produces a quadratic equation in one variable, which is why those are the only three outcomes. Solve it, then use the line to find each partner value.

What you'll learn

Solve a linear-quadratic system by substitution
Solve the same system by graphing and compare the results
Explain why such a system has two, one or no solutions

A line meets a parabola

\begin{cases} y = x^2 - 2x - 3 \\[2pt] y = x + 1 \end{cases}

A solution is a point on both graphs, exactly as with two lines. The difference is that a curve can meet a line more than once.

y = x² − 2x − 3
y = x + 1

A parabola and a line

The graph shows two crossings. Substitution finds them exactly. Both equations give $y$ , so set the expressions equal:

x^2 - 2x - 3 = x + 1

x^2 - 3x - 4 = 0 \quad\Rightarrow\quad (x - 4)(x + 1) = 0

So $x = 4$ or $x = -1$ . Each $x$ still needs its $y$ , and the line supplies it:

x = 4:\; y = 4 + 1 = 5 \qquad x = -1:\; y = -1 + 1 = 0

The solutions are $(4, 5)$ and $(-1, 0)$ , matching the graph.

Why there are two, one or no solutions

Substituting the line into the quadratic always produces a quadratic equation in one variable. A quadratic equation has two real roots, one repeated root, or none — and each root gives one point.

The quadratic after substituting has	The line	Solutions
two real roots	crosses the curve	two points
one repeated root	touches the curve	one point
no real roots	misses the curve	none

The discriminant $b^2 - 4ac$ of that quadratic decides which row applies, before any solving.

y = x²
crosses: y = 2x + 3
touches: y = 2x − 1
misses: y = 2x − 4

Three parallel lines against y = x²

$x^2 = 2x + 3$ gives $x^2 - 2x - 3 = 0$ , discriminant $16$ : two points, $(-1, 1)$ and $(3, 9)$ .
$x^2 = 2x - 1$ gives $x^2 - 2x + 1 = 0$ , discriminant $0$ : one point, $(1, 1)$ .
$x^2 = 2x - 4$ gives $x^2 - 2x + 4 = 0$ , discriminant $-12$ : no points.

The three lines are parallel, so the picture is one line sliding downward. As it slides, the discriminant falls from positive through zero to negative, and the crossings go from two, to one, to none.

A line cannot cross a parabola three times. The substitution never produces anything above degree two, and a degree-two equation has at most two roots.

A line meets a circle

The same method handles a circle. The circle of radius $5$ centered at the origin is $x^2 + y^2 = 25$ .

\begin{cases} x^2 + y^2 = 25 \\[2pt] y = x + 1 \end{cases}

Substitute the line for $y$ :

x^2 + (x + 1)^2 = 25 \quad\Rightarrow\quad 2x^2 + 2x - 24 = 0 \quad\Rightarrow\quad x^2 + x - 12 = 0

(x + 4)(x - 3) = 0 \quad\Rightarrow\quad x = -4 \;\text{ or }\; x = 3

The line gives the partners: $x = 3$ pairs with $y = 4$ , and $x = -4$ pairs with $y = -3$ .

y = x + 1

A circle and a line

Use the line, not the circle, to find each $y$ . The circle allows two values of $y$ for most $x$ : at $x = 3$ both $y = 4$ and $y = -4$ satisfy $x^2 + y^2 = 25$ . Only $(3, 4)$ is on the line.

Worked examples

Common mistakes

Practice problems

Solve $y = x^2$ and $y = x + 2$ .

Answer

$(2, 4)$ and $(-1, 1)$

Full solution

$x^2 = x + 2$ gives $x^2 - x - 2 = 0$ , so $(x - 2)(x + 1) = 0$ .

From the line: $x = 2$ gives $y = 4$ , and $x = -1$ gives $y = 1$ .
Solve $y = x^2 - 4$ and $y = 3x$ .

Answer

$(4, 12)$ and $(-1, -3)$

Full solution

$x^2 - 4 = 3x$ gives $x^2 - 3x - 4 = 0$ , so $(x - 4)(x + 1) = 0$ .

From the line: $x = 4$ gives $y = 12$ , and $x = -1$ gives $y = -3$ .
How many solutions does the system $y = x^2 + 3$ and $y = 2x$ have?

Answer

None.

Full solution

$x^2 + 3 = 2x$ gives $x^2 - 2x + 3 = 0$ . The discriminant is $4 - 12 = -8$ , which is negative, so the line misses the parabola.
Solve $y = x^2 + 1$ and $y = 2x$ .

Answer

$(1, 2)$ , the only solution.

Full solution

$x^2 + 1 = 2x$ gives $x^2 - 2x + 1 = 0$ , which is $(x - 1)^2 = 0$ .

The single root $x = 1$ gives $y = 2$ . The line is tangent to the parabola there.
Solve $x^2 + y^2 = 10$ and $y = x + 2$ .

Answer

$(1, 3)$ and $(-3, -1)$

Full solution

$x^2 + (x + 2)^2 = 10$ gives $2x^2 + 4x - 6 = 0$ , so $x^2 + 2x - 3 = 0$ and $(x + 3)(x - 1) = 0$ .

From the line: $x = 1$ gives $y = 3$ , and $x = -3$ gives $y = -1$ .

Check on the circle: $1 + 9 = 10$ ✓ and $9 + 1 = 10$ ✓
Find the points where the line $y = -3x$ meets the circle $x^2 + y^2 = 3$ .

Hint

After substituting, the equation has no $x$ -term. Solve for $x^2$ directly.

Answer

$\left(\tfrac{\sqrt{30}}{10}, -\tfrac{3\sqrt{30}}{10}\right)$ and $\left(-\tfrac{\sqrt{30}}{10}, \tfrac{3\sqrt{30}}{10}\right)$ , about $(0.55, -1.64)$ and $(-0.55, 1.64)$ .

Full solution

$x^2 + (-3x)^2 = 3$ gives $10x^2 = 3$ , so $x^2 = \tfrac{3}{10}$ and $x = \pm\sqrt{\tfrac{3}{10}} = \pm\tfrac{\sqrt{30}}{10}$ .

The line gives each $y$ as $-3x$ , so the positive $x$ pairs with a negative $y$ , and the negative $x$ with a positive $y$ .
For which value of $k$ does the line $y = 2x$ touch the parabola $y = x^2 + k$ at exactly one point?

Hint

One point means the quadratic has a discriminant of zero.

Answer

$k = 1$

Full solution

$x^2 + k = 2x$ gives $x^2 - 2x + k = 0$ .

One solution needs a zero discriminant: $(-2)^2 - 4(1)(k) = 0$ , so $4 - 4k = 0$ and $k = 1$ .
A ball’s height is $h = -16t^2 + 64t$ feet after $t$ seconds. When is it at $60$ feet?

Answer

At $t = 1.5$ and $t = 2.5$ seconds.

Full solution

$-16t^2 + 64t = 60$ gives $16t^2 - 64t + 60 = 0$ , so $4t^2 - 16t + 15 = 0$ .

$(2t - 3)(2t - 5) = 0$ , so $t = 1.5$ or $t = 2.5$ .
Explain why a line cannot cross a parabola in three points.

Answer

Substitution gives a quadratic equation, which has at most two roots, and each root is one crossing.

Full solution

Replacing $y$ with the line’s expression leaves an equation of degree two in $x$ .

An equation of degree two has at most two solutions, and every crossing point supplies one of them. So there are at most two crossings.
Solving $y = x^2 - 1$ and $y = x + 1$ , Mia finds $x = 2$ and $x = -1$ and writes the answer as the point $(2, -1)$ . Find her error.

Hint

What does each $x$ -value belong to?

Answer

The two $x$ -values belong to two different points. The solutions are $(2, 3)$ and $(-1, 0)$ .

Full solution

Her algebra is correct: $x^2 - 1 = x + 1$ gives $x^2 - x - 2 = 0$ , so $x = 2$ or $x = -1$ .

Each $x$ is the first coordinate of its own crossing point. The line gives the partners: $x = 2$ pairs with $y = 3$ , and $x = -1$ pairs with $y = 0$ .

Mia’s $(2, -1)$ is not on the line, since $2 + 1 = 3$ , not $-1$ .

Frequently asked questions

How do I solve a system with a line and a parabola?

Substitute the line's expression for y into the parabola's equation. That gives a quadratic in x. Solve it, then put each x back into the line to get its y.

How many solutions can a line and a parabola have?

Two, one or none. The substitution produces a quadratic, and a quadratic has two, one or no real roots.

What does one solution mean on the graph?

The line touches the curve at a single point without crossing it. The line is tangent there.

Why use the line to find y?

Each x gives exactly one y on a line. Using a circle's equation instead gives two candidate values of y, and only one of them is paired with that x.

Can a line cross a parabola three times?

No. The substitution leaves an equation of degree two, which cannot have more than two roots.

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

CCSS.MATH.CONTENT.HSA.REI.C.7Reasoning with Equations and InequalitiesSolve a simple system consisting of a linear equation and a quadratic equation in two variables algebraically and graphically.

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