Question

Find the equation of tangent line to the circle x^2+y^2+z^2+5x-7y+2z-8=0,3x-2y+4z+3=0 and the point (-3,5,4)

Answer 1

$x^2+y^2+z^2+5x-7y+2z-8=$

$(x+\frac{5}{2})^2+(y-\frac{7}{2})^2+(z+1)^2=\frac{55}{2}$

This is sphere with center $C (-\frac{5}{2},\frac{7}{2},-1)$ and radius $\sqrt{\frac{55}{2}}.$

Find the equation of the tangent plane to the sphere at the point $P(-3,5,4).$

The radius vector or normal is $CP=\begin{pmatrix} -0.5 \\ 1.5\\ 5 \end{pmatrix}.$

A vector in the plane we seek is $v=\begin{pmatrix} x+3 \\ y-5\\ z-4 \end{pmatrix}.$

So, $\begin{pmatrix} -0.5 \\ 1.5\\ 5 \end{pmatrix}*\begin{pmatrix} x+3 \\ y-5\\ z-4 \end{pmatrix}=0\Rarr$

$-0.5(x+3)+1.5(y-5)+5(z-4)=0$.

Answer:

The equation of the tangent plane is

$1.5y-0.5x+5z-29=0.$

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