If the polar of a point P with respect to a circle of radius r which touches the coordinate axes and lies in the first quadrant is x+2y=4r, then the point P is
Let P be (x1,y1) Now, equation of circle of radius r which touches the coordinate axes and lies in the first quadrant is (x−r)2+(y−r)2=r2 ⇒‌‌x2+y2−2xr−2yr−r2=0 Now, polar of P with respect to the circle x2+y2−2xr−2yr−r2=0 is
But it is given that polar of P with respect to above circle is x+2y=4r ∴‌‌‌
x1−r
1
=‌
y1−r
2
=‌
x1+y1−r
4
⇒x1−r=‌
x1+y1−r
4
and x1−r=‌
y1−r
2
⇒4x1−4r=x1+y1−r and 2x1−2r=y1−r ⇒‌‌3x1−3r=y1 and 2x1−r=y1 ∴‌‌3x1−3r=2x1−r ⇒‌‌x1=2r and y1=2x1−r=4r−r=3r ∴‌‌P is (2r,3r)