Lecture 5 [02 March, 2020] MATH 127 (Prof Farid)

 MATH -127 (Coordinate Geometry)
Chapter-2 
(Pair of Straight Lines)

Question 6: Find the angle between the lines joining the origin to the points of intersection of the line  \(y - 3x = 3 \) with  \(x^2 + 2 xy + 3 y^+ 4 x + 8y - 11 = 0\)  

Solution:   Given that,  \(y - 3x = 3 \tag{6.1}\label{eq:6.1}\) 
and  \(x^2 + 2 xy + 3 y^+ 4 x + 8y - 11 = 0 \tag{6.2}\label{eq:6.2}\)                

From equation (6.1) we get,
 \(\frac{y - 3x}{2}= 1 ~~~~~~~~ \textrm{Here,}~~ h = -1, ~~~a = 7, ~~~ b = -1\)                    

Now we can write, 
 \begin{align*} & x^2 + 2xy + 3y^2 + (4x + 8y)\left(\frac{y - 3x}{2}\right) - 11 \left(\frac{y - 3x }{2}\right)^2 = 0 \\ & 7 x^2 - 2xy - y^2 = 0 \tag{6.3}\label{eq:6.3} \end{align*}     

                   

Let, \(\theta\) be the angle between the lines represented by (6.3), then
  $$ \therefore \tan\theta = \frac{2 \sqrt{h^2 - ab}}{a + b}= \frac{2\sqrt{1 + 7}}{7 - 1}= \frac{2 \sqrt{8}}{6}= \frac{2\sqrt{2}}{3} $$  

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Chapter -3

The Circle

    ♦ Definition: A circle is the locus (set of points) of a point that moves so that the distance of a given point is always equal to a given distance. The given point is the center of the circle and the given distance is its radius.

    Standard form of the equation of a circle:  \((x - h)^2 + (y - k )^2 = a^2 \tag{1}\label{eq:1}\)  
where \((h,k)\) is the center and \(a\) = radius \([a \ge 0]\)

        • Case 1: If \(a = 0\), equation (1) represents a point circle.
        • Case 2: If \((0,0)\) is the center and radius = 1, then  \(x^2 + y^2 = 1 \tag{2}\label{eq:2}\) represents a unit circle.
        • Case 3:  \(x^2 + y^2 = a^2 \tag{3}\label{eq:3}\)  

    ♦ General Form:  \begin{align*} & x^2 + y^2 + 2gx + 2fy + c = 0 \tag{4}\label{eq:4} \\ \implies & (x^2 + 2gx + g^2) + ( y^2 + 2 fy + f^2) = g^2 + f^2 - c \\ Now, & ~~\textrm{Completing the square property, } \\ \implies & (x + g)^2 + (y + f )^2 = (\sqrt{g^2 + g^2 - c})^2 \end{align*}  

   

Comparing it to the standard equation,
        center \((- g, - f)\) and radius = \(\sqrt{g^2 + f^2 - c}\)

    ♦ Polar form of a circle :
•Pole on fixed point or initial point.
•Polar on fixed line or initial line.

  $$ {r_1}^2 - 2 r_1 r_2 \cos ( \theta_1 - \theta) = a^2 $$  

        • Case 1: if the pole lies on the circle,
then, \(r_1 = a\), In this case  \(r = 2a~\cos(\theta_1 - \theta)\)  
        • Case 2: If \(\theta = 0^{\circ}\) the initial line is the diameter and in this case, we have  \(r = 2a~\cos\theta_1\)
        • Case 3: if \(x = r\cos\theta\) and \(y = r \sin\theta\) then  \begin{align*} & x^2 + y^2 = a^2 \\ & \implies r^2 = a^2 \\ & \therefore r = a \end{align*}  

Upon which condition a point lies outside or inside a circle: 
  \begin{align*} & \textrm{radius = }~ \sqrt{g^2 + f^2 - c} ~~~~~~~~~~ \textrm{center = }~ (- g, - f) \\ & ~~~~~ \textrm{if, } d > R \rightarrow \textrm{(outside)} .\\ & ~~~~~ \textrm{if, } d = R \rightarrow \textrm{(on the circle)} \\ & ~~~~~ \textrm{if, } d < R \rightarrow \textrm{(inside)} \end{align*}  

Now, general forms of circle are:  \(x^2 + y^2 = a^2 ~~~~~ \textrm{and}~~~~~ x^2 + y^2 + 2gx + 2fy + c = 0\)  
    

        (1) The equation of the tangent line to the circle \(x^2 + y^2 = a^2 ~ \textrm{at}~ (x_1, y_1) \textrm{is, }~ xx_1 + yy_1 = a^2\)
        (2) The equation of the tangent line to the circle  \(x^2 + y^2 + 2gx + 2fy + c = 0~ \textrm{at}~ (x_1, y_1) ~ \textrm{is, } ~ xx_1 + yy_1 + g ( x + x_1) + f (y + y_1) + c = 0\)  

♦ Some points to know:
    • Pole is a point, Polar is the line.
    • Pole may be inside the circle or outside the circle.
        If the pole is outside, polar will pass through the circle.
        If the pole is inside, polar will be outside the circle.