Fined the equestion of the circle which passes through (4,1) , (-4,-3) and (2,-5)
Answers
Step-by-step explanation:
We can write the equation of the circle in the form:
a
x
+
b
y
+
c
=
x
2
+
y
2
This equation is satisfied by the
(
x
,
y
)
pairs
(
4
,
3
)
,
(
−
2
,
−
5
)
and
(
5
,
2
)
So we have:
⎧
⎪
⎨
⎪
⎩
4
a
+
3
b
+
c
=
25
−
2
a
−
5
b
+
c
=
29
5
a
+
2
b
+
c
=
29
Subtracting the first equation from the second and third to eliminate
c
, we get:
{
−
6
a
−
8
b
=
4
a
−
b
=
4
Subtracting
8
times the second of these equations from the first, we get:
−
14
a
=
−
28
Hence:
⎧
⎪
⎨
⎪
⎩
a
=
2
b
=
−
2
c
=
23
So the equation of our circle can be written:
2
x
−
2
y
+
23
=
x
2
+
y
2
Subtracting the left hand side from the right, we find:
0
=
x
2
−
2
x
+
y
2
+
2
y
−
23
0
=
x
2
−
2
x
+
1
+
y
2
+
2
y
+
1
−
25
0
=
(
x
−
1
)
2
+
(
y
+
1
)
2
−
5
2
So we can write the equation as:
(
x
−
1
)
2
+
(
y
+
1
)
2
=
5
2
This is (more or less) in the form:
(
x
−
h
)
2
+
(
y
−
k
)
2
=
r
2
where
(
h
,
k
)
=
(
1
,
−
1
)
is the centre of the circle and
r
=
5
the radius:
graph{((x-1)^2+(y+1)^2-5^2)((x-1)^2+(y+1)^2-0.02)((x-4)^2+(y-3)^2-0.01)((x-5)^2+(y-2)^2-0.01)((x+2)^2+(y+5)^2-0.01) = 0 [-12, 12, -7, 4.4]}