Further examples on intersections
Question
Show that the line \(y = 2x + 1\) intersects the circle \({x^2} + {y^2} - 6x - 7y + 9 = 0\) and determine the points of intersection.
The method is sustitution:
\(y = 2x + 1\)
\({x^2} + {y^2} - 6x - 7y + 9 = 0\)
\({x^2} + {(2x + 1)^2} - 6x - 7(2x + 1) + 9 = 0\)
Multiply out the brackets and collect terms:
\({x^2} + 4{x^2} + 4x + 1 - 6x - 14x - 7 + 9 = 0\)
\(5{x^2} - 16x + 3 = 0\)
Factorise the quadratic:
\((5x - 1)(x - 3) = 0\)
\(x = \frac{1}{5},3\)
\(y = \frac{7}{5},7\)
Therefore the line \(2x + 1\) intersects the circle at \(\left( {\frac{1}{5},\frac{7}{5}} \right)\) and \((3,7)\).
Question
Show that the line \(3y = 2x - 8\) is a tangent to the circle \({x^2} + {y^2} - 4x - 6y = 0\) and determine the point of contact.
The method is sustitution:
\(3y = 2x - 8\)
\(y = \frac{2}{3}x - \frac{8}{3}\)
\({x^2} + {y^2} - 4x - 6y = 0\)
\({x^2} + {\left( {\frac{2}{3}x - \frac{8}{3}} \right)^2} - 4x - 6\left( {\frac{2}{3}x - \frac{8}{3}} \right) = 0\)
Multiply out the brackets and collect terms:
\({x^2} + \frac{4}{9}{x^2} - \frac{{32}}{9}x + \frac{{64}}{9} - 4x - 4x + 16 = 0\)
\(9{x^2} + 4{x^2} - 32x + 64 - 36x - 36x + 144 = 0\)
\(13{x^2} - 104x + 208 = 0\)
Factorise the quadratic:
\(13({x^2} - 8x + 16) = 13(x - 4)(x - 4) = 0\)
\(x = 4\) ie equal roots. \(y = 0\)
The line touches the circle at \((4,0)\).
Question
Show that the line \(y = 2x - 8\) does not intersect the circle \({x^2} + {y^2} - 2x - 2y - 3 = 0\).
The method is sustitution:
\(y = 2x - 8\)
\({x^2} + {y^2} - 2x - 2y - 3 = 0\)
\({x^2} + {(2x - 8)^2} - 2x - 2(2x - 8) - 3 = 0\)
Multiply out the brackets and collect terms:
\({x^2} + 4{x^2} - 32x + 64 - 2x - 4x + 16 - 3 = 0\)
\(5{x^2} - 38x + 77 = 0\)
Quadratic should not factorise, so find the discriminant:
\({b^2} - 4ac = {( - 38)^2} - 4 \times 5 \times 77 = - 96\)
\({b^2} - 4ac\) is negative, therefore there are no real roots.
Therefore the line \(y = 2x - 8\) misses the circle.
