In Algebra 1 you found where a line meets a parabola. The same method, substitution, finds where any two curves meet, including the circles you met in Geometry.
Key idea
Solve one equation for a variable (or for a chunk like $x^2$) and substitute into the other. Solve the single-variable equation you get, then go back for the other coordinate. Keep every solution, and check each pair in both equations.
Why it works
A solution is a point on both curves, so its coordinates make both equations true at once. Substitution builds that requirement into one equation. The number of solutions matches the picture: a line can miss a circle, touch it once or cross it twice; a circle and a parabola can meet up to four times.
Worked examples
Worked example: A line and a circle
Solve $x^2 + y^2 = 25$ and $y = x - 1$.
- Substitute: $x^2 + (x - 1)^2 = 25$, so $2x^2 - 2x + 1 = 25$.
- $2x^2 - 2x - 24 = 0$, divide by 2: $x^2 - x - 12 = 0$, so $(x - 4)(x + 3) = 0$.
- $x = 4$ gives $y = 3$; $x = -3$ gives $y = -4$.
Answer: $(4, 3)$ and $(-3, -4)$
Worked example: A circle and a parabola
Solve $x^2 + y^2 = 25$ and $y = x^2 - 5$.
- From the parabola, $x^2 = y + 5$. Substitute the chunk: $y + 5 + y^2 = 25$.
- $y^2 + y - 20 = 0$, so $(y + 5)(y - 4) = 0$.
- $y = 4$: $x^2 = 9$, so $x = 3$ or $x = -3$. $y = -5$: $x^2 = 0$, so $x = 0$.
Answer: $(3, 4)$, $(-3, 4)$ and $(0, -5)$
Common mistake
Losing solutions by dividing by a variable. If you reach $x^2 = 3x$ and divide by $x$, you keep $x = 3$ but lose $x = 0$. Instead, move everything to one side and factor: $x(x - 3) = 0$. The same care applies when $x^2 = 9$: there are two values, $3$ and $-3$.