Find an equation of a sphere with center (2,-3,6) that touch (a) the xy-plane, (b) the yz-plane, (c) the xz-plane.

Step-by-step explanation:

To find the equation of a sphere with center (2,-3,6) that touch

(a) the xy-plane,

The equation of xy plane is z=0

Hence the radius would be distance from z=0 perpendicular to xy plane = z coordinate =6

Equation is

(b) the yz-plane,

The equation of yz plane is x=0

Hence the radius would be distance from x=0 perpendicular to yz plane = x coordinate =2

Equation is

(c) the xz-plane.

The equation of zx plane is y=0

Hence the radius would be distance from y=0 perpendicular to zx plane = |y coordinate| =3

Equation is

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?