Equations of straight lines and circles examples I


  • Examples
    (a) If circle C passes through three points P(0, -4), Q(3, 5) and R(3, -5), find the equation of circle C.
    (b) (i) Find the distance between the centre of the circle and A(8, 6).
         (ii) Hence determine whether A lies inside circle C. Explain briefly.
  • Solutions

    (a) Let \({x^2} + {y^2} + Dx + Ey + F = 0\) be the equation of circle C.
    ∵ P(0, -4), Q(3, 5) and R(3, -5) lie on circle C.
    ∴ \(\begin{array}{1}{0^2} + {( - {\rm{ }}4)^2} + D(0) + E( - {\rm{ }}4) + F = 0\\16 - 4E + F = 0\;................\;(1)\\{3^2} + {5^2} + D(3) + E(5) + F = {0^{^{}}}\\34 + 3D + 5E + F = 0\;................\;(2)\\{3^2} + {( - 5)^2} + D(3) + E( - 5) + F = {0^{^{}}}\\34 + 3D - 5E + F = 0\;................\;(3)\end{array}\)
    (2) - (3), \(\begin{array}{1}10E = 0\\E = {0^{}}\end{array}\)
    Substitute \(E = 0\) into (1),
    \(\begin{array}{1}16 - 4(0) + F = 0\\F = - {16^{}}\end{array}\)
    Substitute \(E = 0\) and \(F = - 16\) into (2),
    \(\begin{array}{1}34 + 3D + 5(0) - 16 = 0\\D = - {\rm{ }}6\end{array}\)
    ∴ The equation of circle C is \({x^2} + {y^2} - 6x - 16 = 0\).

    (b) (i) Coordinates of the centre\(\begin{array}{l}
         = ( - \frac{{ - {\rm{ }}6}}{2}{\rm{ }},{\rm{ }}0)\\ = (3{\rm{ }},{\rm{ }}0)\end{array}\)
         Required distance\(\begin{array}{l} = \sqrt {{{(8 - 3)}^2} + {{(6 - 0)}^2}} \\ = \underline{\underline {\sqrt {61} }} \end{array}\)
    (ii) Radius of circle C\(\begin{array}{l}
         = \sqrt {{{(\frac{{ - {\rm{ }}6}}{2})}^2} + {0^2} - ( - 16)} \\ = 5\end{array}\)
          ∵ \(\sqrt {61} > 5\)
          ∴ A does not lie inside circle C.

  • Examples
    In the figure, P is the centre of the circle. If radius \(OP = \sqrt 2 \), find the equation of the circle.

  • Solutions
    Let (x, y) be the coordinates of P.
    \(\begin{array}{1}\sin 45^\circ = \frac{y}{{\sqrt 2 }}\\\frac{{\sqrt 2 }}{2} = \frac{y}{{\sqrt 2 }}\\y = {1^{}}\end{array}\)
    \(\begin{array}{1}\cos 45^\circ = \frac{x}{{\sqrt 2 }}\\\frac{{\sqrt 2 }}{2} = \frac{x}{{\sqrt 2 }}\\x = {1^{}}\end{array}\)
    ∴ The coordinates of P are (1, 1).
    ∴ The equation of the required circle is
    \(\begin{array}{1}{(x - 1)^2} + {(y - 1)^2} = {(\sqrt 2 )^2}\\{x^2} - 2x + 1 + {y^2} - 2y + 1 = {2^{^{}}}\\{x^2} + {y^2} - 2x - 2y = {0^{^{}}}\end{array}\)
  • Examples
    Given two points A(-1, 1) and B(2, 0), P(x, y) is a moving point such that \(P{A^2} + P{B^2} = 20\).
    (a) Express \(P{A^2}\) and \(P{B^2}\) in terms of x and y.
    (b) According to the condition given, write an equation in x and y.
    (c) Is the equation obtained in (b) an equation of a circle? If yes, find the coordinates of the centre and the radius of the circle. (Leave your answers in surd form if necessary.)
  • Solutions
    \(\begin{array}{1} (a) P{A^2} = {[x - ( - 1)]^2} + {(y - 1)^2}\\ = {(x + 1)^2} + {(y - 1)^2}^{^{}}\\ = {\underline{\underline {{x^2} + {y^2} + 2x - 2y + 2}} ^{^{}}}\end{array}\)
    \(\begin{array}{1}P{B^2} = {(x - 2)^2} + {(y - 0)^2}\\ = {\underline{\underline {{x^2} + {y^2} - 4x + 4}} ^{^{}}}\end{array}\)

    \(\begin{array}{1} (b) P{A^2} + P{B^2} = 20\\{x^2} + {y^2} + 2x - 2y + 2 + {x^2} + {y^2} - 4x + 4 = {20^{^{^{}}}}\\2{x^2} + 2{y^2} - 2x - 2y - 14 = {0^{^{^{}}}}\\{x^2} + {y^2} - x - y - 7 = {0^{^{^{}}}}\end{array}\)

    (c) Yes.
    Coordinates of the centre\(\begin{array}{l} = ( - \frac{{ - {\rm{ }}1}}{2},{\rm{ }} - {\rm{ }}\frac{{ - {\rm{ }}1}}{2})\\ = {\underline{\underline {(\frac{1}{2},{\rm{ }}\frac{1}{2})}} ^{^{}}}\end{array}\)
    Radius\(\begin{array}{l} = \sqrt {{{(\frac{{ - {\rm{ }}1}}{2})}^2} + {{(\frac{{ - {\rm{ }}1}}{2})}^2} - ( - 7)} \\ = {\underline{\underline {\sqrt {\frac{{15}}{2}} }} ^{^{}}}\end{array}\)