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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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)

Vedclass · Accepted Answer

$6x-8y+15=0$

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:

Similar Questions

If $A(2, c)$ and $B(d, 2)$ are two points such that the polar of one point with respect to the circle $x^2+y^2=16$ passes through the other,then $c+d=$

Consider the circles $S_1: x^2+y^2+2x+8y-23=0$ and $S_2: x^2+y^2-4x+10y+19=0$. If the polars of the centre of one circle with respect to the other circle are $L_1$ and $L_2$,then $L_1$ and $L_2$ are

The inverse of the point $(1, 2)$ with respect to the circle $x^2 + y^2 - 4x - 6y + 9 = 0$ is

If the inverse of $P(-3, 5)$ with respect to a circle is $(1, 3)$,then the polar of $P$ with respect to that circle is

The polar of the point $(5, -1/2)$ with respect to the circle $(x - 2)^2 + y^2 = 4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module