DSE Core Math Notes
DSE Core Math Notes
DSE Core Math Notes

Equations of straight lines and circles examples II


  • Examples
    Consider two circles \({C_{{\rm{ }}1}}{\kern 1pt} :{x^2} + {y^2} + 2x - 8y + 1 = 0\) and \({C_{{\rm{ }}2}}{\kern 1pt} :{x^2} + {y^2} - 6x - 14y + 57 = 0\).
    (a) Find the coordinates of the centres and the radii of \({C_{{\rm{ }}1}}\) and \({C_{{\rm{ }}2}}\).
    (b) Find the distance between the centres.
    (c) Hence prove that \({C_{{\rm{ }}1}}\) and \({C_{{\rm{ }}2}}\) touch each other.
  • Solutions

    (a) Coordinates of the centre of \({C_{{\rm{ }}1}}\)\(\begin{array}{l}\\ = ( - \frac{2}{2},{\rm{ }} - {\rm{ }}\frac{{ - {\rm{ }}8}}{2})\\ = {\underline{\underline {( - 1{\rm{ }},{\rm{ }}4)}} ^{}}\end{array}\)
    Radius of \({C_{{\rm{ }}1}}\)\(\begin{array}{1}\\ = \sqrt {{{(\frac{2}{2})}^2} + {{(\frac{{ - {\rm{ }}8}}{2})}^2} - 1} \\ = {\underline{\underline 4} ^{}}\end{array}\)
    Coordinates of the centre of \({C_{{\rm{ }}2}}\)\(\begin{array}{l}\\ = ( - \frac{{ - {\rm{ }}6}}{2},{\rm{ }} - {\rm{ }}\frac{{ - {\rm{ }}14}}{2})\\ = {\underline{\underline {(3{\rm{ }},{\rm{ }}7)}} ^{}}\end{array}\)
    Radius of \({C_{{\rm{ }}2}}\)\(\begin{array}{1}\\ = \sqrt {{{(\frac{{ - {\rm{ }}6}}{2})}^2} + {{(\frac{{ - {\rm{ }}14}}{2})}^2} - 57} \\ = \underline{\underline {{\rm{ }}1{\rm{ }}}} _{}^{}\end{array}\)

    (b) Distance between the centres\(\begin{array}{l} = \sqrt {{{[3 - ( - 1)]}^2} + {{(7 - 4)}^2}} \\ = {\sqrt {16 + 9} ^{}}\\ = {\underline{\underline 5} ^{}}\end{array}\)

    (c) ∵ Radius of \({C_{{\rm{ }}1}}\) + Radius of \({C_{{\rm{ }}2}}\)\(\begin{array}{l}\\ = 4 + 1\\ = {5^{}}\end{array}\)
    = Distance between the centres
    ∴ \({C_{{\rm{ }}1}}\) and \({C_{{\rm{ }}2}}\) touch each other.

  • Examples
    G(-1, 0) is the centre of circle C. Straight line \(L{\kern 1pt} :y = - 2x + k\) is the tangent to the circle at P.
    (a) Express the coordinates of P in terms of k.
    (b) If \(x + y + 4 = 0\) passes through P, find the value of k.
    (c) Find the equation of circle C.
  • Solutions
    (a) Slope of L\( = - 2\)
    ∵ L is perpendicular to GP.
    ∴ Slope of GP\( = \frac{1}{2}\)
    The equation of GP is
    \(\begin{array}{1}y - 0 = \frac{1}{2}[x - ( - 1)]\\y = \frac{1}{2}(x + 1)\end{array}\)
    \(\left\{ \begin{array}{l}y = - 2x + k\;................\;(1)\\y = \frac{1}{2}(x + 1)\;...............\;(2)\end{array} \right.\)
    Substitute (2) into (1),
    \(\begin{array}{1}\frac{1}{2}(x + 1) = - 2x + k\\x + 1 = - {\rm{ }}4x + 2k\\5x = 2k - 1\\x = \frac{{2k - 1}}{5}\end{array}\)
    Substitute \(x = \frac{{2k - 1}}{5}\) into (1),
    \(\begin{array}{1}y = - 2(\frac{{2k - 1}}{5}) + k\\ = \frac{{k + 2}}{5}\end{array}\)
    ∴ The coordinates of P are \((\frac{{2k - 1}}{5}{\rm{ }},{\rm{ }}\frac{{k + 2}}{5})\).

    \(\begin{array}{1} (b) \frac{{2k - 1}}{5} + \frac{{k + 2}}{5} + 4 = 0\\2k - 1 + k + 2 + 20 = 0\\3k = - 21\\k = \underline{\underline { - {\rm{ }}7}} \end{array}\)

    (c) The coordinates of P are \((\frac{{2( - 7) - 1}}{5}{\rm{ }},{\rm{ }}\frac{{ - {\rm{ }}7 + 2}}{5})\), i.e. (-3, -1).
    The equation of circle C is
    \(\begin{array}{1}{[x - ( - 1)]^2} + {(y - 0)^2} = {[ - 1 - ( - 3)]^2} + {[0 - ( - 1)]^2}\\{(x + 1)^2} + {y^2} = 5\\{x^2} + 2x + 1 + {y^2} = 5\\{x^2} + {y^2} + 2x - 4 = 0\end{array}\)
  • Examples
    In the figure, a circle with its centre at C is the inscribed circle of ΔOAB, where P, Q and R are the points of contact.


    (a) Find the coordinates of C.
    (b) Find the equation of the circle.
  • Solutions
    (a) Let (x, x) be the coordinates of C.
    \(\begin{array}{1}BR = 6 - x\\BP = 6 - x\end{array}\) (tangents from ext. pt.)
    \(\begin{array}{1}AQ = 8 - x\\AP = 8 - x\end{array}\) (tangents from ext. pt.)
    In ΔOAB,
    \(\begin{array}{1}A{B^2} = O{A^2} + O{B^2}\\ = {8^2} + {6^2}\\ = {100^{}}\\AB = {10^{}}\end{array}\) (Pyth. theorem)
    ∴ \(\begin{array}{1}(6 - x) + (8 - x) = 10\\14 - 2x = {10^{}}\\x = {2^{}}\end{array}\)
    ∴ The coordinates of C are (2, 2).

    (b) The equation of the circle is
    \(\begin{array}{1}{(x - 2)^2} + {(y - 2)^2} = {2^2}\\{x^2} - 4x + 4 + {y^2} - 4y + 4 = 4\\{x^2} + {y^2} - 4x - 4y + 4 = 0\end{array}\)