11made the correction... it was 'curves'
(x-1)2 + (y-2)2=32
&& (x-5)2 - (y-6)2=-63
Find the circle of min. radius which cuts the curves orthogonally.
This is quite a difficult problem !!
-
UP 0 DOWN 0 2 9
9 Answers
62take a circle of the form
x2+y2+2gx+2fy+c=0
then use the condition of orthogonality..
1consider the line 3x+4y=12 , intersecting x and y axes at points A and B . Let C be the feet of the perpendicular drawn from O (origin) upon AB . Again D And E be the feet of the perpendicular drawn from C on OA and OB respectively. if r is inradius and R is circumradius find the value of
r \bigtriangleup abc +r\bigtriangleup OEC+R\bigtriangleup BEC
11@ Nishant sir dunno circle nahin hai
2nd waala curve kya hai ( hope it is a hyperbola )
Please clear it
11@ mani-- ya 2nd eqn is hyperbola.
@nishant sir- im getting lost with too many unknowns.
62(x-1)2 + (y-2)2=9
the normal to this circle is any line passing through the center.
so in general it is
(x-1)+λ(y-2)=0
x+λy-1-2λ=0
now this should be tangential to the 2nd curve.. for the point where both the curves intersect
62I went mad above :P
Seee the solution of the above that i gave and use the same method..
I mean you will haev to introduce a new circle and then do the whole thing..
But I think you can do it this way!