Where a Line Meets a Circle

Substitute, solve, and send each x back.

Where, exactly?

The circle x² + y² = 25 has its center at the origin and a radius of 5: every point on it is 5 from (0, 0). The straight line y = x + 1 runs across it. A drawing shows that the line crosses the circle twice, but a point read off a drawing is only an estimate. Algebra finds the two points exactly.

xyy = x + 1

The line y = x + 1 crosses the circle x² + y² = 25 in two places, one in the top right and one in the bottom left.

One equation in x

A point where the line meets the circle lies on both of them, so its coordinates make both equations true at once. The two equations are a pair of simultaneous equations, one linear and one quadratic.

The line says that y is equal to x + 1. So wherever the circle's equation has y, write x + 1 instead: x² + (x + 1)² = 25. The y has gone, and what is left is one equation in x alone.

Expand and gather

Expand the bracket first. (x + 1)² means (x + 1)(x + 1), which is x² + 2x + 1, so the equation becomes x² + x² + 2x + 1 = 25.

Collect the two x² terms: 2x² + 2x + 1 = 25. Subtract 25 from both sides so that one side is 0: 2x² + 2x − 24 = 0. Every coefficient is even, so divide both sides by 2: x² + x − 12 = 0. That is an ordinary quadratic equation.

Take care with the bracket. (x + 1)² is not x² + 1, because the middle term 2x comes from multiplying x by 1 twice. Without it the equation would be 2x² + 1 = 25, whose roots are x = √12 and x = −√12, and neither of those is where the line meets the circle.

Solve, then find each y

Factor the quadratic. Two numbers that multiply to −12 and add to 1 are 4 and −3, so (x + 4)(x − 3) = 0. One of the brackets must be 0, so x = −4 or x = 3.

Each value of x belongs to one crossing, and each crossing needs its y. Put each x into the equation of the line, y = x + 1. When x = −4, y = −4 + 1 = −3. When x = 3, y = 3 + 1 = 4.

Use the line for this step, not the circle. Putting x = 3 into the circle gives 9 + y² = 25, so y² = 16 and y = 4 or y = −4. Both of those points are on the circle, but only (3, 4) is on the line, because 3 + 1 is 4 and not −4.

xyy = x + 1(−4, −3)(3, 4)

The two solutions are the two crossing points, (−4, −3) and (3, 4).

Check both points

Each point must satisfy both equations. For (−4, −3): (−4)² + (−3)² = 16 + 9 = 25, and −4 + 1 = −3. For (3, 4): 3² + 4² = 9 + 16 = 25, and 3 + 1 = 4. Both points are on the circle and on the line.

The part of the line between the two points is a chord of the circle: a straight line segment whose two ends are on the circle.

Two crossings, one or none

A line can cross a circle twice, touch it once, or miss it altogether. The quadratic says which, because each real root is a crossing.

Horizontal lines show this most simply. Put y = 3 into x² + y² = 25: x² + 9 = 25, so x² = 16 and x = 4 or x = −4. The line y = 3 crosses the circle twice, at (−4, 3) and (4, 3). Notice that x² = 25 − 9 is the difference of the squares. Subtracting the lengths, 5 − 3 = 2, does not give the crossing.

Put y = 5: x² + 25 = 25, so x² = 0 and x = 0. The two roots have become one repeated root, and the line touches the circle at the single point (0, 5). A line that touches a circle at exactly one point is a tangent. Put y = 7: x² + 49 = 25, so x² = −24. No real number squares to a negative, so the line y = 7 passes above the circle without meeting it.

For a sloping line, the discriminant b² − 4ac of the quadratic decides. The line y = x + 1 gave x² + x − 12 = 0, with discriminant 1² − 4 × 1 × (−12) = 1 + 48 = 49. That is positive, so there are two crossings. The line y = x + 8 gives x² + (x + 8)² = 25, which becomes 2x² + 16x + 39 = 0. Its discriminant is 16² − 4 × 2 × 39 = 256 − 312 = −56. That is negative, so there are no real roots and the line misses the circle.

xyy = 3y = 5

The line y = 3 crosses the circle at (−4, 3) and (4, 3). The line y = 5 touches it only at (0, 5), the top of the circle, so it is a tangent.

Inside or outside the circle

A point is inside the circle x² + y² = r² when its distance from the center is less than r. By Pythagoras that distance is √(x² + y²), so compare x² + y² with r². If x² + y² is less than r², the point is inside; if it is equal, the point is on the circle; if it is more, the point is outside. The point (2, 3) gives 4 + 9 = 13, which is less than 25, so it is inside the circle x² + y² = 25.

Worked example: The Range of a Radio Mast, a Town and a Straight Road

Question A radio mast stands at the origin of a map on which 1 unit is 1 km, with x measured to the east and y to the north. The signal reaches every place within 10 km of the mast. (a) A town is at (5, −8). Does the town receive the signal? (b) A straight road follows the line y = x + 2. Find the coordinates of the points where the road enters and leaves the range of the mast.

  1. 1.A point (x, y) is √x2 + y2 km from the mast. On the edge of the range this distance is 10 km, so the edge is the circle x2 + y2 = 102, which is x2 + y2 = 100.

    −10−50510−10−50510km east of the mast, xkm north, ymast10 kmevery point on the edge is 10 km from the mastx2+ y2= 102, so x2+ y2= 100
    −10−50510−10−50510km east of the mast, xkm north, ymast10 kmevery point on the edge is 10 km from the mastx2+ y2= 102, so x2+ y2= 100
    Every point on the edge of the range is 10 km from the mast at the origin, so the edge is the circle x2 + y2 = 100.

  2. 2.For the town at (5, −8), x2 + y2 = 52 + (−8)2 = 25 + 64 = 89. (a) Yes. Since 89 < 100, the town is less than 10 km from the mast, so it is inside the circle and receives the signal.

    −10−50510−10−50510km east of the mast, xkm north, ymasttown (5, −8)10 kmtown: 52+ (−8)2= 25 + 64 = 8989 < 100: the town is inside the circle
    −10−50510−10−50510km east of the mast, xkm north, ymasttown (5, −8)10 kmtown: 52+ (−8)2= 25 + 64 = 8989 < 100: the town is inside the circle
    (a) For the town, 52 + (−8)2 = 25 + 64 = 89. This is less than 100, so the town is inside the circle and receives the signal.

  3. 3.Where the road meets the edge, both equations hold. Substitute y = x + 2 into the circle: x2 + (x + 2)2 = 100. Expand: 2x2 + 4x + 4 = 100. Subtract 100 from both sides and divide both sides by 2: x2 + 2x − 48 = 0.

    −10−50510−10−50510km east of the mast, xkm north, yy = x + 2masttown (5, −8)on the road y = x + 2: x2+ (x + 2)2= 1002x2+ 4x + 4 = 100x2+ 2x − 48 = 0
    −10−50510−10−50510km east of the mast, xkm north, yy = x + 2masttown (5, −8)on the road y = x + 2: x2+ (x + 2)2= 1002x2+ 4x + 4 = 100x2+ 2x − 48 = 0
    Substitute y = x + 2 into the circle: x2 + (x + 2)2 = 100. Expand and divide both sides by 2: x2 + 2x − 48 = 0.

  4. 4.Factorize: two numbers with a product of −48 and a sum of 2 are 8 and −6, so (x + 8)(x − 6) = 0, and x = −8 or x = 6. Substitute each root into y = x + 2: y = −6 when x = −8, and y = 8 when x = 6.

    −10−50510−10−50510km east of the mast, xkm north, ymasttown(−8, −6)(6, 8)(x + 8)(x − 6) = 0, so x = −8 or x = 6y = −8 + 2 = −6, and y = 6 + 2 = 8
    −10−50510−10−50510km east of the mast, xkm north, ymasttown(−8, −6)(6, 8)(x + 8)(x − 6) = 0, so x = −8 or x = 6y = −8 + 2 = −6, and y = 6 + 2 = 8
    Factorize: (x + 8)(x − 6) = 0, so x = −8 or x = 6. The line gives y = −6 and y = 8.

  5. 5.(b) The road enters the range at (−8, −6) and leaves it at (6, 8). Check in the circle: (−8)2 + (−6)2 = 64 + 36 = 100 and 62 + 82 = 36 + 64 = 100.

    −10−50510−10−50510km east of the mast, xkm north, ymasttown(−8, −6)(6, 8)the road enters at (−8, −6) and leaves at (6, 8)check: 64 + 36 = 100, and 36 + 64 = 100
    −10−50510−10−50510km east of the mast, xkm north, ymasttown(−8, −6)(6, 8)the road enters at (−8, −6) and leaves at (6, 8)check: 64 + 36 = 100, and 36 + 64 = 100
    (b) The road enters the range at (−8, −6) and leaves it at (6, 8). Both points satisfy x2 + y2 = 100.

Answer: (a) Yes: 52 + (−8)2 = 89, which is less than 100; (b) at (−8, −6) and (6, 8)

Common mistakes

  • Comparing 89 with 10 and deciding that the town is out of range. The number 89 is the square of the distance, so it must be compared with 102 = 100. The distance itself is √89 ≈ 9.4 km.
  • Expanding (x + 2)2 as x2 + 4. The middle term is missing: (x + 2)2 = x2 + 4x + 4. Without the 4x the quadratic equation has different roots, and the points found are not on the road.

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