Pope
Lv 7
Pope asked in Science & MathematicsMathematics · 10 years ago

Locus of centers of orthogonal spheres?

Two spheres are given:

x² + y² + z² + 14x − 4y − 147 = 0

x² + y² + z² + 4x − 28y - 168z + 6631 = 0

Write a single equation representing the locus of centers of spheres that are orthogonal to both of the given spheres.

2 Answers

Relevance
  • Duke
    Lv 7
    10 years ago
    Favourite answer

    Subtract both equations and simplify to get the required locus:

    5x + 12y + 84z - 3389 = 0.

    This is an equation of a plane, perpendicular to the line, connecting both centers

    Z' (-7, 2, 0) and Z" (-2, 14, 84),

    containing the radical axes of any 2 circles - intersections of both spheres with arbitrary plane through the line, connecting the centers (follow the link below to read a Wiki Article on the subject):

    http://en.wikipedia.org/wiki/Radical_axis

    This plane is a locus of points, having the same power with respect to both spheres (read 'Algebraic construction' Section in the article to get an idea why the equation of the radical axis is obtained by subtracting the equations of both circles - the same approach works in 3D space too).

  • Anonymous
    4 years ago

    I even have reached an identical end as gianlino worked out as below. enable (a, b) be the middle and r1 = radius of the 1st circle and (c, d) be the middle and r2 = radius of the 2nd circle. enable (h, ok) be the middle and r = radius of the variable circle which touches the above 2 circles. => distance of the middle of the variable circle from the middle of the 1st given circle = r + r1 and distance of the middle of the variable circle from the middle of the 2nd given circle = r + r2 => ?[(h - a)^2 + (ok - b)^2] = r + r1 and ?[(h - c)^2 + (ok - d)^2] = r + r2 Subtracting, l ?[(h - a)^2 + (ok - b)^2] - ?[(h - c)^2 + (ok - d)^2] l = l r1 - r2 l changing (h, ok) by (x, y) supplies the locus as l ?[(x - a)^2 + (y - b)^2] - ?[(x - c)^2 + (y - d)^2] l = l r1 - r2 l => Any element P (x, y) on the locus of the curve is such that the constructive super difference of its distances from the facilities of the given circles = constructive super difference of the lengths of the radii of the circles. => the curve is a hyperbola whose foci are the facilities of the given circles and the size of whose transverse axis = (a million/2) l r1 - r2 l. Edit: in reality, from the hyperbola, a factor is lacking mendacity between the intersecting factors of the given 2 circles. So the set of things on the locus does not variety the excellent hyperbola, yet they lie on the hyperbola given by the above equation. additionally, the standards on the locus will lie in ordinary terms on one in all the two arms of the hyperbola and not the two. Edit: contained in right here Wolfram Alpha link, 2 circles with unequal radii and the locus of the facilities of the circles tangent to the two is shown. The factors on one arm of the hyperbola interior the link exterior the two circles variety the locus. Edit: gianlino I study your feedback. I had the circles touching given circles externally in innovations. using fact the asker has not reported it, I agree we could consistently evaluate circles touching the given circles internally as properly as touching one internally and the different externally. pondering this, i will visualize that the standards on hyperbola between the intersecting factors gets lined. in spite of the shown fact that, i could not see how the 2nd branch of hyperbola additionally would be a piece of the locus.

Still have questions? Get answers by asking now.