Question

Find the centre and radius of the circle x^2 + y^2 + z^2 + 2x + 2y + 4z = 3, 2x - y - z = 3.

Answer 1

Answer on Question #45094 – Math – Analytic Geometry

Task:

Find the centre and radius of the circle $x^2 + y^2 + z^2 + 2x + 2y + 4z = 3$, $2x - y - z = 3$.

Solution:

The equation $x^2 + y^2 + z^2 + 2x + 2y + 4z = 3$ can be rewritten in the form

$(x+1)^2 + (y+1)^2 + (z+2)^2 = 9$. As we can see this is a sphere with centre at $(-1, -1, -2)$ and radius 3.

Since the circle lies in the plane $2x - y - z = 3$. Then the vector $(2, -1, -1)$ is perpendicular to the circle. The line that passes through the centre of the sphere and has directing vector $(2, -1, -1)$ is $(x+1)/2 = (y+1)/(-1) = (z+2)/(-1)$. So, the centre of the circle is a point that is in the intersection of the line $(x+1)/2 = (y+1)/(-1) = (z+2)/(-1)$ and of the plane $2x-y-z=3$.

Solving this we get a point $x = 1/4$, $y = -3/4$, $z = -7/4$.

Also it can be easily seen that the point $x = 0$, $y = -3$, $z = 0$ belongs to the circle.

So the radius of the circle is the distance between points $(0, -3, 0)$ and $(1/4, -3/4, -7/4)$ and it is equal to 1.92.

Answer: 1.92.

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