Equations of straight lines and circles  (12.112.4)
Understand the possible intersection of two straight lines

Theory
\(\begin {cases}x+2y5=0\\ xy=0\end{cases}\)
\(\begin{cases} x=\frac{5}{3} \\ y=\frac{5}{3} \end{cases}\)

Examples

Solutions
Understand the equation of a circle

Theory

ExamplesGiven two points A(4,1) and B(4, 7), find the equation of the circle with AB as a diameter.

SolutionsCoordinates of the centre
=Coordinates of the midpoint 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
\(\begin {cases}x+2y5=0\\ x^2+y^2+xy9=0\end{cases}\)
\(\begin{cases} x=\frac{2\sqrt{109}}{5} \qquad y=\frac{23+\sqrt{109}}{10} \\ x=\frac{2+\sqrt{109}}{5} \qquad y=\frac{23\sqrt{109}}{10} \end{cases}\)

ExamplesFind 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).