Sphere

Author Bhawna asked on 2019-07-22 13:26:00

Let  C be the  centre  of  the required  sphere  which touches the plane 4x+3y-47=0 at the point P(8,5,4)

CP is Perpendicular to tangent Plane

Direction Ratio of CP  are 4,3,0

Equation of  CP   are

(x-8)/4=(y-5)/3=(z-4)/0=r

Any Point on CP  is  C(4r+8,3r+5,4)

Radius of sphere = CP = Under root{ (4r+8-8)2+(3r+5-5)2+0}=Under  root {16r2+9r2} =5r= C

Equation of Given Sphere is

X2+y2+z2=1 , centre (0,0,0), Radius =1=C’

CC’=Radius of First Sphere +Radius of Second Sphere

CC’=5r+1

Under root {(4r+8)2+(3r+5)2+4=5r+1

Squaring both and  solve it we will get

r= -26/21  and r=-1

taking r= -26/21 

C(4r+8,3r+5,4) = C{4(-26/21)+8,3(-26/21)+5,4) =C(64/21,-27/21,4)

Now taking r= -1

C(4r+8,3r+5,4) = C(-4+8,-3+5,4) = C(4,2,4)