The polar of a point with respect to the circ | vedclass

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(D) The given equation of the circle is $x^2+y^2-10x+12y-3=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we have $g=-5, f=6, c=-3$. The center is $(-g, -f)

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$6x-8y+15=0$

Vedclass · Answer

(D) The given equation of the circle is $x^2+y^2-10x+12y-3=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we have $g=-5, f=6, c=-3$. The center is $(-g, -f) = (5, -6)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{25+36+3} = \sqrt{64} = 8$. $A$ line $Ax+By+C=0$ is the polar of a point $(x_1, y_1)$ with respect to the circle $x^2+y^2+2gx+2fy+c=0$ if $\frac{x_1+g}{A} = \frac{y_1+f}{B} = \frac{gx_1+fy_1+c}{-C}$. Testing option $(d)$ $6x-8y+15=0$,where $A=6, B=-8, C=15$: $\frac{x_1-5}{6} = \frac{y_1+6}{-8} = \frac{-5x_1+6y_1-3}{-15}$. Solving this system,we find the pole $(x_1, y_1)$ lies inside the circle because the distance from the center $(5, -6)$ to the line $6x-8y+15=0$ is $d = \frac{|6(5)-8(-6)+15|}{\sqrt{6^2+(-8)^2}} = \frac{|30+48+15|}{10} = \frac{93}{10} = 9.3$. Since $d > r$ $(9.3 > 8)$,the line $6x-8y+15=0$ is a line outside the circle,which is a valid polar for a point inside the circle.

The polar of a point with respect to the circle $x^2+y^2-10x+12y-3=0$,which is neither a tangent nor a chord of contact,is:

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