Equations of straight lines and circles examples II

ExamplesConsider 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.

ExamplesG(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
\(\begin{array}{1}{[x  (  1)]^2} + {(y  0)^2} = {[  1  (  3)]^2} + {[0  (  1)]^2}\\{(x + 1)^2} + {y^2} = 5\\{x^2} + 2x + 1 + {y^2} = 5\\{x^2} + {y^2} + 2x  4 = 0\end{array}\)

ExamplesIn 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
\(\begin{array}{1}{(x  2)^2} + {(y  2)^2} = {2^2}\\{x^2}  4x + 4 + {y^2}  4y + 4 = 4\\{x^2} + {y^2}  4x  4y + 4 = 0\end{array}\)