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Determine the equation of a sphere that is tangent to the plane $x+y+z-9=0$ at the point $M(2,3,4)$ and whose center belongs to the plane: $$ (P) : 7x-4y+5z-14=0 $$ My solution: it is clear that the point $A(2,5,4)$ belongs to the plane $(P)$. Then, let $d$ be the line passing through $A$, $d$ orthogonal to $(P)$ (so that the director vector is equal with the normal vector of the plane, $v(7,-4,5)$ ). From the parametric equations of this line I found that the center of the sphere is of the form $C(2+7t, 5-4t, 4+5t)$. At this point, if I calculate the distance from the center of the sphere to the first plane I get ${|8t+11|\over\sqrt{3}}=r$, where $r$ is the radius of the sphere. And I'm stuck at this point because I don't really see how I can use the point $M$ to solve this problem. Any ideas? Thanks!

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3 Answers 3

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The line orthogonal to the plane $x+y+z-9=0$ and passing through $M$ is the line$$\{(2,3,4)+\lambda(1,1,1)\mid\lambda\in\Bbb R\}.\tag1$$The center of the sphere must belong to $(1)$ and also to the plane $P$. So, let us compute the point at which they intersect. In order to do that, one solves the equation$$7(2+\lambda)-4(3+\lambda)+5(4+\lambda)-14=0;$$its only solution is $\lambda=-1$. So, the sphere is centered at $(1,2,3)$ and its radius is the distance from $(1,2,3)$ to $(2,3,4)$, which is $\sqrt{3}$. So, it's the sphere described by$$(x-1)^2+(y-2)^2+(z-3)^2=3.$$

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Consider the sphere with center at $M(2,3,4)$ and radius zero $$(x-2)^2+(y-3)^2+(z-4)^2=0$$ A linear combination of that degenerate sphere and the tangent plane $P:x+y+z-9=0$ represents any sphere tangent to $P$ at the point $M$.

$$(x-2)^2+(y-3)^2+(z-4)^2+\lambda(x+y+z-9)=0\tag{1}$$ expland and collect $$x^2+y^2+z^2+(\lambda -4) x+(\lambda -6) y+(\lambda -8) z+29-9 \lambda =0$$ whose center is $$C_\lambda=\left(\frac{4-\lambda}{2},\frac{6-\lambda}{2},\frac{8-\lambda}{2}\right)$$ in order to belong to the plane $7x-4y+5z-14=0$ we must have $$\frac{7 (4-\lambda )}{2}-2 (6-\lambda )+\frac{5 (8-\lambda )}{2}-14=0$$ which gives $\lambda=2$. Substitute this value in the equation $(1)$ to get $$x^2+y^2+z^2-2 x-4 y-6 z+11=0$$

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It is possible to work backwards. Consider the equation of a sphere:

$$F(x,y,z) = (x-a)^2 + (y-b)^2 + (z-c)^2$$ $$\nabla F = \left<2(x-a), 2(y-b), 2(z-c)\right>\Rightarrow \nabla(2,3,4) = \left<2(2-a), 2(3-b), 2(4-c)\right>$$

so the tangent plane is thus $2(2-a)(x-2) + 2(3-b)(y-3) + 2(4-c)(z-4) = 0$. Comparing coefficients with $x+y+z - 9 = 0$, the centre is at $a=1, b=2, c=3$, which does belong to the given plane.

Thus the squared distance from the centre $(1,2,3)$ to $(2,3,4)$ is $3$, which yields $(x-1)^2+(y-2)^2+(z-3)^2=3$.

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