Answer to Question #89742 in Analytic Geometry for Anjali

Question #89742
Find the equation of the sphere which pass through the circle x^2+y^2+z^2=9,2x+2y-7=0& the touch the plane x-y+z+3=3
1
Expert's answer
2019-05-15T08:43:21-0400

The sphere


"x^2+y^2+z^2-9+\\lambda(2x+2y-7)=0"

passes through the given circle for all values of "\\lambda."

Its center is "(-\\lambda,\\ -\\lambda,\\ 0 )."

Radius is "\\sqrt{2\\lambda^2+7\\lambda+9}."

If a plane "x-y+z+3=0" touches a sphere then the lenght of the the perpendicular from its centre to the plane is equal to its radius

"{|-\\lambda+\\lambda+0+3| \\over \\sqrt{(1)^2+(-1)^2+(1)^2}}=\\sqrt{2\\lambda^2+7\\lambda+9}"

"2\\lambda^2+7\\lambda+9=3"

"2\\lambda^2+7\\lambda+6=0"

"\\lambda={ -7\\pm\\sqrt{(7)^2-4(2)(6)}\\over 2(2)}={ -7\\pm1\\over 4}""\\lambda_1=-2,\\ \\lambda_2=-{3 \\over 2}"

"\\lambda_1=-2"

Its center is "(2,\\ 2,\\ 0 )."

Radius is "\\sqrt{2(-2)^2+7(-2)+9}=\\sqrt{3}."

The equation of the sphere is


"(x-2)^2+(y-2)^2+z^2=3"

"\\lambda_2=-{3 \\over 2}"

Its center is "({3 \\over 2},\\ {3 \\over 2},\\ 0)."

Radius is "\\sqrt{2(-{3 \\over 2})^2+7(-{3 \\over 2})+9}=\\sqrt{3}."

The equation of the sphere is


"(x-{3 \\over 2})^2+(y-{3 \\over 2})^2+z^2=3"

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS