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.
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1So a circle in 3-dimensions? – Simply Beautiful Art Jan 23 '16 at 12:44
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A curve in$\mathbf R^3$ requires at least $2$ equations. – Bernard Jan 23 '16 at 13:00
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Hint: By symmetry, the center is of the form ${\lambda,\lambda,\lambda}$. The three points (hence the circle) live in the plane $x+y+z=3$, so you can read the center off. – lulu Jan 23 '16 at 13:01
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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.
Simply Beautiful Art
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I tohught that equation above is an equation of a sphere,tha'ts why I was confused – userasadfg Jan 23 '16 at 12:58
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Oh, you ask that? A circle can only exist between three points that are coplanar because a circle is 2-dimensional. If you did find the equation though, it would still be a sphere. – Simply Beautiful Art Jan 23 '16 at 13:01
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@user307370 consider the the sphere equation where $z=0$. This produces the circle equation. – Simply Beautiful Art Jan 23 '16 at 13:02
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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 $\Omega(1,1,1)$.
Bernard
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You sure it goes along the lines $x^2+y^2+z^2=9$? Doesn't the center of the sphere have any effect? – Simply Beautiful Art Jan 23 '16 at 13:10
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There are other spheres that will intersect the plane in the circle circumscribed to triangle $MNP$, i.e. the system of equations that define the circle is not unique. The system of equations I propose is the simplest, and anyway, the centres of the spheres are all on the line $O\Omega$. – Bernard Jan 23 '16 at 13:17