Intersection of a line and circle
There are three ways a line and a circle can be associated, ie the line cuts the circle at two distinct points, the line is a tangent to the circle or the line misses the circle.
To work out which case you have, use algebra to work out how many points of intersection there are.
- If the line cuts through the circle, there will be two points of intersection
- If the line is a tangent to the circle, there will be one point of intersection
- If the line misses the circle, there will be no point of intersection
Example
The diagram below shows the circle \({x^2} + {y^2} + 18x + 20y + 81 = 0\) and three lines:
- \(y = x + 1\) which appears to cut the circle in two points
- \(x = 1\) which appears to be a tangent to the circle
- \(y = - x + 3\) which appears to miss the circle
Here's the algebra which should confirm our observations.
The method is substitution.
\(y = x + 1\)
\({x^2} + {y^2} + 18x + 20y + 81 = 0\)
\({x^2} + {(x + 1)^2} + 18x + 20(x + 1) + 81 = 0\)
Multiply out the brackets and collect terms.
\({x^2} + {x^2} + 2x + 1 + 18x + 20x + 20 + 81 = 0\)
\(2{x^2} + 40x + 102 = 0\)
Factorise the quadratic:
\(2({x^2} + 20x + 51) = 0\)
\(2(x + 3)(x + 17) = 0\)
Therefore \(x = - 3, - 17\)and \(y = - 2, - 16\)
Therefore the line \(y = x + 1\) intersects the circle at \(( - 3, - 2)\) and \(( - 17, - 16)\).
Again using the substitution method.
\(x = 1\)
\({x^2} + {y^2} + 18x + 20y + 81 = 0\)
\({1^2} + {y^2} + 18(1) + 20y + 81 = 0\)
Multiply out the brackets and collect terms.
\(1 + {y^2} + 18 + 20y + 81 = 0\)
\({y^2} + 20y + 100 = 0\)
Factorise the quadratic:
\((y + 10)(y + 10) = 0\)
\(y = - 10\)
\(x = 1\)
Therefore the line \(x = 1\) intersects the circle at \((1, - 10)\). So it is a tangent.
Again using the substitution method.
\(y = - x + 3\)
\({x^2} + {y^2} + 18x + 20y + 81 = 0\)
\({x^2} + {( - x + 3)^2} + 18x + 20( - x + 3) + 81 = 0\)
Multiply out the brackets and collect terms.
\({x^2} + {x^2} - 6x + 9 + 18x - 20x + 60 + 81 = 0\)
\(2{x^2} - 8x + 150 = 0\)
Quadratic does not factorise fully, so find the discriminant:
\(2({x^2} - 4x + 75) = 0\)
\({b^2} - 4ac = {( - 4)^2} - 4 \times 1 \times 75 = - 284\)
\({b^2} - 4ac\) is negative, therefore there are no real roots.
Therefore the line \(y = - x + 3\) misses the circle.
