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How to find the tangent plane to sphere say, $x^{2} + y^{2} + z^{2} +2ux+2vy+2wz+d=0$ from the point $P(x1,y1,z1)$ not necessarily on the sphere.

If possible, solve using 3D not vectors.

user1611542
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  • Care to explain the downvote ? – user1611542 Aug 20 '16 at 17:09
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    Link it to "What has been so far done?" – Narasimham Aug 20 '16 at 17:09
  • I tried calculating for a given sphere and point. First finding the intersection point of plane and sphere (Assuming l,m,n as direction cosines). Also finding one more equation by making perpendicular distance equal to radius. But it did not seem to be the right approach as l,m,n could not be easily solved from it. – user1611542 Aug 20 '16 at 17:11
  • Solutions on the web used partial derivatives and vectors, which I am not allowed to use right now! – user1611542 Aug 20 '16 at 17:12
  • You cannot use vectors, why? – Narasimham Aug 20 '16 at 17:17
  • The topic is not in my syllabus. – user1611542 Aug 20 '16 at 17:18
  • There is a problem; there is an infinite number of planes issued from a given point and tangent to the sphere.... – Jean Marie Aug 20 '16 at 17:23
  • Really ? How ? Shouldn't there be max 2 such planes. – user1611542 Aug 20 '16 at 17:29
  • You have a point,you have the centre of the sphere, thus you have a vector which is perpendicular to the plane and (from above) a point on the plane thus you can describe the plane. – Meadara Aug 20 '16 at 18:02
  • About the fact there are an infinite number of planes tangent to a sphere S and passing through an exterior point P : do you agree with me that from P one can draw tangent lines to S that generate a cone. Along each of these lines L, one can draw a tangent plane to the cone which is as well a tangent plane to the sphere ? – Jean Marie Aug 20 '16 at 21:30
  • Oh yeah, I got the point. – user1611542 Aug 21 '16 at 12:09

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Hint: You can calculate a second point, which would satisfy the equation of a plane. Find the second point by using the vectors: $\mathbf{x}',\mathbf{y'}, \mathbf{z'}$. $$\mathbf{x}'=\dfrac{\mathbf{x}}{\sqrt{\mathbf{x}^2+\mathbf{y}^2+\mathbf{z}^2}}, \mathbf{y}'=\dfrac{\mathbf{y}}{\sqrt{\mathbf{x}^2+\mathbf{y}^2+\mathbf{z}^2}}, \mathbf{z}'=\dfrac{\mathbf{z}}{\sqrt{\mathbf{x}^2+\mathbf{y}^2+\mathbf{z}^2}}$$

Each of this vectors should satisfy the following equation:

$$(x_1+u)^2+(y_1+v)^2+(z_1+w)^2-[\mathbf{x'}(x_1+u)+\mathbf{y'}(y_1+v)+\mathbf{z'}(z_1+w)]^2=r^2$$

Applying this to the sphere $x^{2} + y^{2} + z^{2} +2ux+2vy+2wz+d=0$ gives: $$x_1^2+y_1^2+z_1^2+2x_1u+2y_1v+2z_1w+d-[\mathbf{x'}(x_1+u)+\mathbf{y'}(y_1+v)+\mathbf{z'}(z_1+w)]^2=0$$

Curious
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