Question

Find the equations of two tangent planes to the sphere x2+y2+z2=9 which pass through the line ,x+y=6 ,x-2z=3

Answer 1

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Answer 2

Concept

The perpendicular distance from the center of the sphere to the plane should be equal to the radius of the sphere, i.e. d=R, where d is the perpendicular distance and R is the radius of the sphere.

Given

The equation of the sphere,

$x^{2} + y^{2} + z^{2} = 9$, hence radius R = 3

And the two equations of lines through which the tangent plane passes is given as,

$x+y=6\\x-2z=3$

Find

We are  asked to calculate the tangent plane of the given sphere.

Solution

We can write,

$(x+y-6)+k(x-2z-3)=0\\x(1+k)+y-2kz-(6+3k)=0$

The formula to calculate the perpendicular distance is given as,

$d = \frac{ax+by+cz+d}{\sqrt{a^{2} + b^{2} + c^{2} } }$

where the points x, y, z will be the center of the sphere. Since in this case the center is at origin, therefore

$d =modulus \frac{-(6+3k)}{\sqrt{(1+k)^{2} + 1^{2} + (-2k)^{2} } }\\d = \frac{(6+3k)}{\sqrt{(1+k)^{2} + 1^{2} + (-2k)^{2} } } = R\\\frac{(6+3k)}{\sqrt{(1+k)^{2} + 1^{2} + (-2k)^{2} } } = 3$

After solving the above equation, we get

$k = 1, k= -1/2$

Putting these values of k into the above equation we get the required equations of the tangent plane is

$2x+y-2z=0\\x+2y+2z=9$

Hence the equations of the tangent plane are 2x+y-2z=0 and x+2y+2z=9.

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