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 {16r^{2}+9r^{2}} =5r= C

Equation of Given Sphere is

X^{2}+y^{2}+z^{2}=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^{2 }=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)

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 {16r^{2}+9r^{2}} =5r= CEquation of Given Sphere is

X

^{2}+y^{2}+z^{2}=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^{2 }=5r+1Squaring 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)