Vikas TU
Last Activity: 7 Years ago
Dear Student,
given: it must be xsin alpha+ycos alpha=p intersects x^2+y^2=a^2
solving these 2 equations we get:
x=±sqrt(a^2-p^2) and y=-p
and theta=2n*pi+pi and n belongs to Z
eqn of circle having (x1,y1) and( x2,y2) as diameter ends:
(x-x1)(x-x2)+(y-y1)(y-y2)=0 –(1)
so here: they are (sqrt(a^2-p^2),-p) and (-sqrt(a^2-p^2),-p)
putting in eqn (1):
(x- sqrt(a^2-p^2))(x+ sqrt(a^2-p^2))+(y+p)(y+p)=0
therefore, x^2+y^2-2py+2p^2-a^2=0 is eqn of reqd circle.[ans]
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)
