Question

A plane which is tangent to the sphere $(x-a{)}^2+(y-a{)}^2+(z-a{)}^2=2a^2$
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 $x-2a=\frac{y-a}{-2}=z-0$
Then the equation of the tangent plane is

• A. $x-z=2a$
• B. $x+z=2a$
• C. $-x+z=2a$
• D. $-x-z=2a$

Answer 1

The correct option is A $x-z=2a$

Let equation of the plane which makes equal intercepts to coordinate axes is $x+y+z=k$
$\because$ it passes through $(a,a,a)$
$\therefore x+y+z=3a$
Let the required equation of the plane is $l(x-2a)+m(y-a)+n(z-0)=0\dots \left(1\right)$
where $l,m,n$ are the direction cosines
distance of the $(a,a,a)$ from the $\left(1\right)$
$|-al+0+an|=a|-l+n|=1\dots \left(2\right)$
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
$(t+a),(-2t),(t-a)$
$\therefore t+a+4t+t-a=0\Rightarrow t=0$
$\therefore P(2a,a,0)$
So distance is $=2a$
$\because$ line is also tangent to the sphere
$\therefore$ both the planes are perpendicular
$\therefore l+m+n=0\dots \left(3\right)$
and $l-2m+n=0\dots \left(4\right)$
Solving the above equations
$x-z=2a$