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Question

# A plane which is tangent to the sphere (x−a)2+(y−a)2+(z−a)2=2a2 a normal is drawn to the sphere which makes equal intercepts to the coordinate axis. Let the the equation of the line of the intersection is x−2a=y−a−2=z−0 Then the equation of the tangent plane is

A
xz=2a
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B
x+z=2a
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C
x+z=2a
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D
xz=2a
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Solution

## The correct option is A x−z=2aLet equation of the plane which makes equal intercepts to coordinate axes is x+y+z=k ∵ it passes through (a,a,a) ∴x+y+z=3a Let the required equation of the plane is l(x−2a)+m(y−a)+n(z−0)=0…(1) where l,m,n are the direction cosines distance of the (a,a,a) from the (1) |−al+0+an|=a|−l+n|=1…(2) distance of the line from the center of the sphere is any general point on the line is P(t+2a,−2t+a,t) Direction ratios of the line perpendicular to given line is (t+a),(−2t),(t−a) ∴t+a+4t+t−a=0⇒t=0 ∴P(2a,a,0) So distance is =2a ∵ line is also tangent to the sphere ∴ both the planes are perpendicular ∴l+m+n=0…(3) and l−2m+n=0…(4) Solving the above equations x−z=2a

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