Finding equation of a circle

Question

I am supposed to find an equation of a circle passing though points M(3,0,0),N(0,3,0) and P(0,0,3) I have tried to find an intersection of bisectors of lines MN and NP but ended up with nothing.

Answer 1

It's not as hard as you may believe, in fact, you would use the following equation:

$$(x-a)^2+(y-b)^2+(z-c)^2=r^2$$

Just an altered version of the circle equation where $C(a,b,c)$ is the center of our circle.

So all you do is plug in you numbers to get 3 equations:

$$(3-a)^2+b^2+c^2=r^2\tag{1}$$

$$a^2+(3-b)^2+c^2=r^2\tag{2}$$

$$a^2+b^2+(3-c)^2=r^2\tag{3}$$

Subtract $(1)$ and $(2)$ to get:

$$(3-a)^2-a^2+b^2-(3-b)^2=0$$

$$9-6a-9+6b=0$$

$$6b=6a\to a=b$$

Subtract $(2)$ and $(3)$:

$$(3-b)^2-b^2+c^2-(3-c)^2=0$$

Following from above steps, we get:

$$a=b=c$$

From this, I want to assume that $a=b=c=0$ because I think it is obvious with the three points you chose and I want to have $r=3$, such that we have:

$$x^2+y^2+z^2=3^2=9$$

Plugging in your points say that this works and we have the following:

$$3=\pm\sqrt{x^2+y^2+z^2}=\pm\sqrt{x^2+(\sqrt{y^2+z^2})^2}$$

Basically, this is applying Pythagorean Theorem twice so that the points $x_1,y_1,z_1$ are all equidistant to the origin.

Since ask specifically for the circle, here you go:

$$(x-1)^2+(y-1)^2+(z-1)^2=6$$

Just remember to use the following equation: $x+y+z=3$ to maintain the fact that it is a circle.

Answer 2

Such a circle is the intersection of the sphere centred at origin that passes through $M,N,P$ and the plane through these points. Hence it is defined by the equations $$\begin{cases} x+y+z=3,\\ x^2+y^2+z^2=9. \end{cases}$$ The centre of the circle is the projection of the origin onto the plane, i.e. the point $\Omega(1,1,1)$.