Where curves meet
Tonight's idea
Solving a system of equations is asking a geometric question: where do these graphs meet? Every solution \((x, y)\) is a point that sits on both curves at once. Two lines meet once, never (parallel), or everywhere (the same line twice). With a curve involved, one method handles every case.
Substitution. Solve one equation for a variable and feed it into the other. One equation in one unknown comes out. For a line meeting a circle or a parabola, each root fixes exactly one point, so the quadratic's discriminant \(b^2 - 4ac\) counts the meeting points: positive means two, zero means one (the curves just touch, a tangent), negative means they miss. Otherwise a root is not a point: back-substitute and count the real points you get (the stretch has two roots and four points).
Elimination. When both equations are linear, or share a matching term, add or subtract multiples of them so a variable disappears. For two circles, subtracting the equations removes both \(x^2\) and \(y^2\) at once and leaves a line, and the meeting points must lie on it.
Watch one: a line through a circle
Where does \(y = x + 1\) meet \(x^2 + y^2 = 25\)? Substitute: \(x^2 + (x+1)^2 = 25\), so \(x^2 + x - 12 = 0\), which factors as \((x + 4)(x - 3) = 0\). Then \(x = 3\) gives \(y = 4\), and \(x = -4\) gives \(y = -3\). Check both in the circle: \(9 + 16 = 25\) and \(16 + 9 = 25\). Two points: \((3, 4)\) and \((-4, -3)\). Slide the line up to \(y = x + 8\) and substitution gives \(2x^2 + 16x + 39 = 0\), discriminant \(256 - 312 \lt 0\): the line misses.
Your turn
- Find where the line \(y = 2x - 1\) meets the parabola \(y = x^2 - 4x + 7\).
- For which value of \(c\) does the line \(y = 2x + c\) touch the parabola \(y = x^2\) at exactly one point? Find that point.
- Two circles: \(x^2 + y^2 = 25\) and \((x - 6)^2 + y^2 = 13\). Subtract one equation from the other (expand \((x - 6)^2\) first), solve for \(x\), then find both meeting points.
- Where does the parabola \(y = x^2\) meet the circle \(x^2 + y^2 = 6\)? Substitute \(x^2 = y\). You will get two values of \(y\). Explain why one of them must be thrown away.
Stretch: how many times can two circles meet?
Argue that two different circles can meet at most twice. (Hint: subtracting the equations gives a line. How many times can a line meet a circle?) Then find all the points where \(y = x^2\) meets \(x^2 + (y - 3)^2 = 5\). A parabola and a circle can do better than two.
Answers
- \(x^2 - 4x + 7 = 2x - 1\), so \(x^2 - 6x + 8 = 0\), \((x-2)(x-4) = 0\). Points \((2, 3)\) and \((4, 7)\).
- \(x^2 = 2x + c\) gives \(x^2 - 2x - c = 0\), discriminant \(4 + 4c\). Zero when \(c = -1\). Then \(x^2 - 2x + 1 = 0\), \(x = 1\), point \((1, 1)\).
- Subtracting: \(x^2 - (x-6)^2 = 12\), so \(12x - 36 = 12\), \(x = 4\). Then \(y^2 = 9\), \(y = \pm 3\). Points \((4, 3)\) and \((4, -3)\). Check in the second circle: \(4 + 9 = 13\).
- \(y + y^2 = 6\), so \((y + 3)(y - 2) = 0\), \(y = 2\) or \(y = -3\). But \(y = x^2\) can never be negative, so \(y = -3\) gives no real \(x\). With \(y = 2\), \(x = \pm\sqrt2\). Two points: \((\pm\sqrt2, 2)\).
Stretch. Subtracting two circle equations removes \(x^2\) and \(y^2\) and leaves a linear equation, a line. (If the circles share a centre, what is left is a false statement, and they never meet.) Every common point lies on that line, and a line meets a circle at most twice, so two circles meet at most twice. For the parabola and circle, substitute \(x^2 = y\): \(y + (y-3)^2 = 5\), so \(y^2 - 5y + 4 = 0\), \(y = 1\) or \(4\), giving \(x = \pm1\) or \(\pm2\). Four points: \((\pm1, 1)\) and \((\pm2, 4)\). A positive discriminant promised two roots, not two points: each \(y\) here gives two values of \(x\), while in exercise 4 one root gave none.