Question
lf the line xcos α+ ysin α=p and the circle x2+y2=a2 intersect at A and B then the equation of the circle on AB as diameter (x2+y2−a2)+k(xcosα+ysinα−p)=0 then k=
lf the line xcos α+ ysin α=p and the circle x2+y2=a2 intersect at A and B then the equation of the circle on AB as diameter (x2+y2−a2)+k(xcosα+ysinα−p)=0 then k=
Open in App
Solution
The correct option is B −2p
Centre of the circle having AB as diameter is the foot of perpendicular from (0,0) to x cos α+y sin α=p then
x−0cos α=y−0sin α=−(−p)1
x=p cos α and y=p sin α
Equation of circle having centre (p cos α,p sin α) and
radius(r)=√a2−(√p2(sin2α+cos2α))2
=√a2−p2
∴(x−p cos α)2+(y−p sinα)2=(√a2−p2)2
x2+y2−a2+p2−x2p cos α−2py sin α+p2=0
(x2+y2−a2)−2p(2cos α x+2sin α y−p)=0
So, k=−2p
∴AB=2r
Centre of the circle having AB as diameter is the foot of perpendicular from (0,0) to x cos α+y sin α=p then
x−0cos α=y−0sin α=−(−p)1
x=p cos α and y=p sin α
Equation of circle having centre (p cos α,p sin α) and
radius(r)=√a2−(√p2(sin2α+cos2α))2
=√a2−p2
∴(x−p cos α)2+(y−p sinα)2=(√a2−p2)2
x2+y2−a2+p2−x2p cos α−2py sin α+p2=0
(x2+y2−a2)−2p(2cos α x+2sin α y−p)=0
So, k=−2p
∴AB=2r
Suggest Corrections
0
Similar questions
View More
Explore more
