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, 

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