# The equation of a sphere tangent to a plane at a point whose center belongs to a plane

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!

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$$(x-1)^2+(y-2)^2+(z-3)^2=3.$$

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 $$x^2+y^2+z^2-2 x-4 y-6 z+11=0$$

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 $$(x-1)^2+(y-2)^2+(z-3)^2=3$$.