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Intersection Of Line And Circle

Intersection of a line and circle

At that place are three ways a line and a circumvolve 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 piece of 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 exist two points of intersection
  • If the line is a tangent to the circle, there volition be one bespeak of intersection
  • If the line misses the circle, there will be no point of intersection

Example

The diagram below shows the circle \({10^2} + {y^two} + 18x + 20y + 81 = 0\) and three lines:

  1. \(y = 10 + 1\) which appears to cut the circumvolve in two points
  2. \(x = 1\) which appears to be a tangent to the circumvolve
  3. \(y = - x + 3\) which appears to miss the circle
Line y=x+1, line x=1 and line y=-x+3 all touch the same point. Line y=x+1 cuts the circle in two points, x=1 is a tangent to the circle, while y=-x+3 misses the circle completely

Here'southward the algebra which should ostend our observations.

The method is substitution.

\[y = x + 1\]

\[{x^2} + {y^2} + 18x + 20y + 81 = 0\]

\[{10^2} + {(x + i)^2} + 18x + 20(10 + 1) + 81 = 0\]

Multiply out the brackets and collect terms.

\[{x^2} + {x^2} + 2x + one + 18x + 20x + 20 + 81 = 0\]

\[2{x^two} + 40x + 102 = 0\]

Factorise the quadratic:

\[2({x^2} + 20x + 51) = 0\]

\[2(x + iii)(x + 17) = 0\]

Therefore \(x = - three, - 17\) and \(y = - 2, - 16\)

Therefore the line \(y = x + 1\) intersects the circle at \(( - 3, - 2)\) and \(( - 17, - sixteen)\) .

Over again using the substitution method.

\[x = ane\]

\[{10^2} + {y^2} + 18x + 20y + 81 = 0\]

\[{1^two} + {y^2} + 18(one) + 20y + 81 = 0\]

Multiply out the brackets and collect terms.

\[1 + {y^two} + 18 + 20y + 81 = 0\]

\[{y^ii} + 20y + 100 = 0\]

Factorise the quadratic:

\[(y + 10)(y + x) = 0\]

\[y = - 10\]

\[x = 1\]

Therefore the line \(x = 1\) intersects the circle at \((1, - ten)\) . So it is a tangent.

Once more using the substitution method.

\[y = - x + three\]

\[{ten^2} + {y^2} + 18x + 20y + 81 = 0\]

\[{x^2} + {( - x + 3)^2} + 18x + 20( - ten + three) + 81 = 0\]

Multiply out the brackets and collect terms.

\[{x^two} + {x^2} - 6x + 9 + 18x - 20x + 60 + 81 = 0\]

\[two{x^two} - 8x + 150 = 0\]

Quadratic does not factorise fully, then find the discriminant:

\[two({x^2} - 4x + 75) = 0\]

\[{b^2} - 4ac = {( - 4)^2} - 4 \times 1 \times 75 = - 284\]

\({b^2} - 4ac\) is negative, therefore there are no real roots.

Therefore the line \(y = - x + 3\) misses the circumvolve.

Intersection Of Line And Circle,

Source: https://www.bbc.co.uk/bitesize/guides/z9pssbk/revision/5

Posted by: dustinhatiltas.blogspot.com

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