Question

find the distance of the centre of the circle x^2+y^2+z^2+x-2y+2z=3, 2x+y+2z=1 from the plane ax+by +cz=d, where a,b,c,d are constants

Answer 1

S
→
x
2
+
y
2
+
z
2
+
x
−
2
y
+
2
z
−
3
=
0
and the plane
Π
1
→
2
x
+
2
z
−
3
=
0
or
Π
1
→
⟨
p
−
p
1
,
→
n
⟩
=
0
with
p
=
(
x
,
y
,
z
)

p
1
=
(
0
,
0
,
3
2
)

→
n
=
(
2
,
2
,
0
)
The circle
C
→
S
∩
Π
1
is centered at the orthogonal projection of the center of S over
Π
1
but
S
→
(
x
+
1
2
)
2
+
(
y
−
1
)
2
+
(
z
+
1
)
2
=
21
4
has as center the point
p
0
=
(
−
1
2
,
1
,
−
1
)
now calling
p
2
the projection of
p
0
onto
Π
1
we know that
p
2
is at the intersection
Π
1
∩
L
with
L
→
p
=
p
0
+
λ
→
n
or
⟨
p
0
−
p
1
+
λ
→
n
,
→
n
⟩
=
0
giving
λ
=
−
⟨
p
0
−
p
1
,
→
n
⟩
∥
∥
→
n
∥
∥
2
and
p
2
=
p
0
−
⟨
p
0
−
p
1
,
→
n
⟩
∥
∥
→
n
∥
∥
2
→
n
=
(
−
3
4
,
3
4
,
−
1
)
now given
Π
2
→
a
x
+
b
y
+
c
z
−
d
=
0
or
Π
2
→
⟨
p
−
p
3
,
→
v
⟩
=
0
with
→
v
=
(
a
,
b
,
c
)

p
3
=
(
0
,
0
,
d
c
)
we have
p
4
as the projection of
p
2
onto
Π
2
as
p
4
=
p
2
−
⟨
p
2
−
p
3
,
→
v
⟩
∥
∥
→
v
∥
∥
2
→
v
and the distance
d
=
∥
p
2
−
p
4
∥
is
d
=
1
4
(
3
a
−
3
b
+
4
(
c
+
d
)
√
a
2
+
b
2
+
c
2