The pole of the line 2x + 3y = 4 with respect | vedclass

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(A) Let the pole be $(x_1, y_1)$. The equation of the polar of the point $(x_1, y_1)$ with respect to the circle $x^2 + y^2 = a^2$ is given by $xx_1 +

Vedclass · Accepted Answer

$(32, 48)$

Vedclass · Answer

(A) Let the pole be $(x_1, y_1)$. The equation of the polar of the point $(x_1, y_1)$ with respect to the circle $x^2 + y^2 = a^2$ is given by $xx_1 + yy_1 = a^2$.Here,$a^2 = 64$,so the equation of the polar is $xx_1 + yy_1 = 64$ $... (i)$.Given that the line $2x + 3y = 4$ is the same as the polar line $xx_1 + yy_1 = 64$,we can write the equation as $xx_1 + yy_1 - 64 = 0$ and $2x + 3y - 4 = 0$.Comparing the coefficients of the two equations:$\frac{x_1}{2} = \frac{y_1}{3} = \frac{-64}{-4}$$\frac{x_1}{2} = 16 \Rightarrow x_1 = 32$$\frac{y_1}{3} = 16 \Rightarrow y_1 = 48$Thus,the pole is $(32, 48)$.

The pole of the line $2x + 3y = 4$ with respect to the circle $x^2 + y^2 = 64$ is:

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