Lecture 6 [09 March, 2020] MATH 127 (Prof Farid)

 

 MATH -127 (Coordinate Geometry)

Chapter -3

The Circle

• In this figure, "O" is the pole and "OP" is the polar ( can be tangent to the circle)

• Length and midpoint of the chord intercepted by  \(x^2 + y^2 = a^2 \tag{1}\label{eq:1}\) on the line  \(y = mx + c \tag{2}\label{eq:2}\)  

Midpoint  \(\left(\frac{- mc}{1 + m^2}, \frac{c}{1 + m^2}\right)\) \begin{align*} & AB = 2 AC = 2 \sqrt{\left\{\frac{a^2 (1 + m^2) - c^2}{1 + m^2}\right\}} \\ & OC \rightarrow ~ y = \frac{- 1}{m}x \\ & ~~~~~~~~~~~\therefore~ x + my = 0 \tag{3}\label{eq:3} \end{align*}  

Since, the equation of AB is,  \(y = mx + c\) , the equation of the perpendicular OC is  \( y = - \frac{1}{m}x \) . Solving (2) and (3) we will get the point C. The length of OA is \(a\) (since radius). Thus we can find A.

The equation of the polar with respect to the circle is  \(x^2 + y^2 = a^2 ~\textrm{is }~ xx_1 + yy_1 = a^2 ~\textrm{at}~ (x_1, y_1)\)  

Question 1: Find the pole of the straight line  \(2x - y = 6\) with respect to the circle  \( 5 x^2 + 5 y^2 = 9\) ( ---------- )

Solution: Given equations are,

  \(2x - y = 6 \tag{1.1}\label{eq:1.1}\) \( 5 x^2 + 5 y^2 = 9 \tag{1.2}\label{eq:1.2}\)  

 Let \(P(x_1,y_1\) be the pole, then the equation of the pole with respect to the circle is given by  \begin{align*} & xx_1 + yy_1 = a^2 \\ \implies & 5xx_1 + 5 yy_1 = 9 \tag{1.3}\label{eq:1.3} \end{align*}  
    Since (1.2) & (1.3) is identical, we can write  \begin{align*} \frac{5x_1 }{2} = \frac{5 y_1}{- 1} = \frac{9}{6} \end{align*}  
    \(\therefore\) the coordinates of the pole are  \(\left(\frac{3}{5}, \frac{- 3}{ 10}\right)\)

    ♦ Combined equation of the pair of tangents: 
  \begin{align*} & x^2 + y^2 = a^2 \\ \implies & x^2 + y^2 - a^2 = 0 \tag{5}\label{eq:5}\\ \textrm{And,} \\ & x^2 + y^2 + 2 gx + 2fy + c = 0 \\ \implies & S = 0 \tag{6}\label{eq:6} \end{align*} \begin{align*} \textrm{for Equation (5),} & \\ & SS_1 = T^2\\ \implies & (x^2 + y^2 - a^2) ({x_1}^2 + {y_1}^2 - a^2) = (ss_1 + yy_1 - a^2)^2\\ \textrm{for Equation (6),} & \\ & SS_1 = T^2 \\ \implies & ( x^2 + y^2 + 2 gx + 2fy + c).({x_1}^2+ {y_1}^2 + 2g x_1 + 2fy_1 + c) \\ & ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \left\{xx_1 + yy_1 + g(x + x_1) + f ( y + y_1) + c \right\} \end{align*}  

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    Question 2: Find the equation of the pair of the tangent drawn from \((0,0)\) to  \( x^2 + y^2 + 2gx + 2fy + c = 0\)  
Solution: The equation of the pair of tangents drawn from \((0,0)\) to  \( x^2 + y^2 + 2gx + 2fy + c = 0\)  

  \begin{align*} & SS_1 = T^2 \\ \implies & (x^2 + y^2 + 2gx + 2fy + c ). c = \left\{x\times 0 + y \times 0 + g(x + 0) + f(y + 0) + c \right\}^2 \\ \implies & c.(x^2 + y^2 + 2gx + 2 fy + c) = ( gx + fy + c)^2 \\ \implies & cx^2 + c y^2 + 2cgx + 2cfy + c^2 = g^2 x^2 + f^2 g^2 + c^2 + 2gfxy + 2fy + 2gxc\\ \implies & c x^2 + c y^2 - g^2 x^2 - f^2 y^2 - 2gxfy = 0 \\ \implies & c(x^2 + y^2) - ( gx + fy )^2 = 0 \\ \implies & C.(x^2 + y^2 ) = (gx + fy )^2 \end{align*}  

    Question 3: Prove that, if the polar of a point "P" with respect to the circle  \( x^2 + y^2 = 37\) touchhes the circle  \(( x - 3 )^2 + (y + 2 )^2 = 2\) . The locas of "P" must be conic.

Solution: From (1) we get, 

  \begin{align*} & xx_1 + yy_1 = 37 \\ & \\ & \frac{|3x_1 - 2y_1 - 37|}{\sqrt{{x_1}^2 + {y_1}^2}} \\ \implies & (3 x_1 - 2 y_1 - 37 )^2 = 25({x_1}^2 + {y_1}^2) \\ \implies & 25 {x_1}^2 + 25 {y_1}^2 + ( - 3 x_1 + 2 y_1 + 37)^2 = 0 \\ \implies & 25 {x_1}^2 + 25 {y_1}^2 + 9 {x_1}^2 + 4 {y_1 }^2 + 1369 - 12 x_1 y_1 + 148y_1 - 111 x_1 = 0 \\ \implies & 34{x_1 }^2 + 29{y_1}^2 - 111 x_1 + 148 y_1 + 1369 = 0 \end{align*}