Equations of straight lines and circles - (12.1-12.4)

Understand the possible intersection of two straight lines

  • Theory
    \(A_1x+B_1x+C_1=0\)
    \(A_2x+B_2x+C_2=0\)

    (Integers only)

    \(\begin {cases}x+2y-5=0\\ x-y=0\end{cases}\)

    \(\begin{cases} x=\frac{5}{3} \\ y=\frac{5}{3} \end{cases}\)

  • Examples
  • Solutions

Understand the equation of a circle

  • Theory
  • Examples
    Given two points A(-4,-1) and B(4, 7), find the equation of the circle with AB as a diameter.
  • Solutions

    Coordinates of the centre
    =Coordinates of the mid-point of AB
    \(\begin{array}{l} = (\frac{{ - {\rm{ }}4 + 4}}{2},{\rm{ }}\frac{{ - 1 + 7}}{2})\\ = (0{\rm{ }},{\rm{ }}3)\end{array}\)
    Radius\(\begin{array}{l} = \frac{1}{2}AB\\ = \frac{1}{2}{\sqrt {{{( - 4 - 4)}^2} + {{( - 1 - 7)}^2}} ^{}}\\ = {\frac{{\sqrt {128} }}{2}^{}}\end{array}\)
    ∴ the equation of the circle is
    \(\begin{array}{c}{(x - 0)^2} + {(y - 3)^2} = {(\frac{{\sqrt {128} }}{2})^2}\\{x^2} + {y^2} - 6y + 9 = 32\\{x^2} + {y^2} - 6y - 23 = 0\end{array}\)
    Alternative method:
    Let P(x, y) be any point on the circle apart from A and B.
    ∵B is a diameter of the circle.
    ∴\(\angle APB = 90^\circ \), i.e. \(AP \bot PB\)
    ∴\(\frac{{y - ( - 1)}}{{x - ( - 4)}} \times \frac{{y - 7}}{{x - 4}} = - 1\)
    ∴The equation of the circle is
    \(\begin{array}{c}(y + 1)(y - 7) = - (x + 4)(x - 4)\\{y^2} - 6y - 7 = - {x^2} + 16\\{x^2} + {y^2} - 6y - 23 = 0\end{array}\)

Find the coordinates of the intersections of a straight line and a circle and understand the possible intersection of a straight line and a circle

  • Theory
    \(A_1x+B_1x+C_1=0\)
    \(x^2+y^2+Dx+Ey+F=0\)

    (Integers only)

    \(\begin {cases}x+2y-5=0\\ x^2+y^2+x-y-9=0\end{cases}\)

    \(\begin{cases} x=\frac{2-\sqrt{109}}{5} \qquad y=\frac{23+\sqrt{109}}{10} \\ x=\frac{2+\sqrt{109}}{5} \qquad y=\frac{23-\sqrt{109}}{10} \end{cases}\)

  • Examples
    Find the coordinates of the points of intersection of circle \(C{\kern 1pt} :{x^2} + {y^2} - 10x + 4y - 3 = 0\)
    and straight line \(L{\kern 1pt} :x - y = 7\).
  • Solutions

    \(\left\{ \begin{array}{l}{x^2} + {y^2} - 10x + 4y - 3 = 0\;.................\;(1)\\x - y = 7\;............................................\;{(2)^{}}\end{array} \right.\)
    From (2), \(x = y + 7\;............................\;(3)\)
    Substitute (3) into (1),
    \(\begin{array}{c}{(y + 7)^2}\, + {y^2}\, - 10(y + 7) + 4y - 3 = 0\\{y^2}\, + 14y + 49 + {y^2}\, - 10y - 70 + 4y - 3 = 0\\2{y^2}\, + 8y - 24 = 0\\{y^2}\, + 4y - 12 = 0\\(y - 2)(y + 6) = {0^{}}\end{array}\)
    \(y = 2\) or \(y = - {\rm{ }}6\)
    Substitute \(y = 2\) into (3), \(\begin{array}{c}\\x = 2 + 7\\ = 9\end{array}\)
    Substitute \(y = - {\rm{ }}6\) into (3), \(\begin{array}{c}\\x = - 6 + 7\\ = 1\end{array}\)
    ∴ The coordinates of the points of intersection are (1,-6) and (9, 2).