find centre and radius of circle in the sphere cut by plane x^2+y^2+z^2-6x-4y+12z=36 , x+2y-2z=1?

Answer 1

i will get the solution geometrically ... not algebraically...

the sphere is
x^2 - 6x + y^2 - 4y + z^2 + 12z = 36
(x²-6x+9) + (y²-4y+4) + (z²+12z+36) = 36 + 9 + 4 + 36
(x-3)² + (y-2)² + (z+6)² = 85

the center is (3,2,-6) with radius √85

the distance of the plane to the center is
|3(1)+2(2)-2(-6) -1|/(3) = 18/3 = 6

using pythagorean theorem... the radius of the desired circle is
r² = 85 - (36) = 49
r = 7


to find the center, we need the vector form... the normal vector of the plane is 〈 1 , 2 , -2 〉
the unit vector is (1/3)〈 1 , 2 , -2 〉

the vector with magnitude 6 is
(2)〈 1 , 2 , -2 〉

we need to add/subtract that to the center of the sphere...
(3,2,-6) - (2 , 4 ,-4) = (1 , -2 , -2)

note that (1,-2,-2) is on the plane...

conclusion
you have your circle with center on (1,-2,-2) and radius 7
§

Answer 2

Find centre and radius of circle in the sphere
x² + y² + z² - 6x - 4y + 12z = 36
cut by the plane x + 2y - 2z = 1.

Rewrite the equation of the sphere is standard form to find the center.

(x² - 6x) + (y² - 4y) + (z² + 12z) = 36
(x² - 6x + 9) + (y² - 4y + 4) + (z² + 12z + 36) = 36 + 9 + 4 + 36
(x - 3)² + (y - 2)² + (z + 6)² = 85

The center of the sphere is P(3, 2, -6).
The radius of the sphere is √85.

The line passing thru the center of the sphere perpendicular to the plane will pass thru the center of the circle in the sphere cut by the plane. The normal vector of the plane is the directional vector of the line.

v = <1, 2, -2>

Now write the equation of the line with directional vector v and point P(3, 2, -6).

L(t) = P + tv
L(t) = <3, 2, -6> + t<1, 2, -2>
L(t) = <3+t, 2+2t, -6-2t>
where t is a scalar ranging over the real numbers

Now substitute these values into the equation of the plane and solve for t.

x + 2y - 2z = 1
(3 + t) + 2(2 + 2t) - 2(-6 - 2t) = 1
3 + t + 4 + 4t + 12 + 4t = 1
9t + 19 = 1
9t = -18
t = -2

x = 3 + t = 3 - 2 = 1
y = 2 + 2t = 2 - 4 = -2
z = -6 - 2t = - 6 + 4 = -2

The center of the circle is Q(1, -2, -2).

The distance from the center of the sphere to the center of the circle is the distance between the two points P and Q.

d² = (3 - 1)² + (2 + 2)² + (-6 + 2)² = 2² + 4² + 4²
d² = 4 + 16 + 16 = 36
d = 6

Let
r = radius of the circle

By the Pythagorean Theorem:

r² = 85 - 6² = 85 - 36 = 49
r = 7

The center of the circle is Q(1, -2, -2) and its radius is 7.

Answer 3

that's particularly straightforward. The circle has those proprieties (x-x0)^2+(y-y0)^2=r^2, r being the radius and X0 and y0 the coordinates of the middle of the circle. x^2-12x+y^2-4y+24=x^2-12x+36+y^2-4y+4-... + (y-2)^2-sixty 4 =>x^2-12x+y^2-4y+24=0<=>(x-6)^2 + (y-2)^2 - sixty 4=0<=>(x-6)^2 + (y-2)^2=sixty 4 => (x-6)^2 + (y-2)^2=8^2. So the middle of the circle is A of coordinates (6,2) and the radius is 8.

Answer 4

simple,
from given equation,
circle lies at center because,x^2=y^2=z^2=same value.so
in the circle of genaral formula x^2+y^2+z^2+2ux+2vy+2hz=k
formula for center=(-u,-v,-h)
so divide coeff of x,y,z by 2 and invert the sign
so center is -6/2=-3=3(invert sign)
similarly
center is(3,2,-6)
for radius
x^2+y^2=r^2
so r=6(by taking sqrt of 36)

Answer 5

It is

24x1y200z64M
OKay?

Answer 6

say what?