Assuming you are using the normal x-y Cartesian coordinates, the standard form for a circle centered on a point, (0, 0)
(
x
0
,
y
0
)
, is: (−0)2 + (−0)2 = 2
(
x
−
x
0
)
2
+
(
y
−
y
0
)
2
=
r
2
.
If the circle is centered on the origin, this becomes the familiar equation: x^2+y^2=r^2 .
Since a diameter is two radii, r= 5 units in your question.
So, the net result with 0 = −4
x
0
=
−
4
, and 0 = 9
y
0
=
9
, is:
( + 4)2 + ( − 9)2 = 52 = 25(1) (1) (x + 4
)
2
+ (y − 9
)
2
=
5
2
= 25
Often we see other forms of the same equation, say, with y as a function of x,
= ±25 − ( +4)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√+9= 9 ± 25 − (2 + 8 + 16)‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√= 9 ± 9 − 8 −2‾‾‾‾‾‾‾‾‾‾‾‾‾√ y = ±
25
−
(
x
+
4
)
2
+9 = 9 ±
25
−
(
x
2
+
8
x
+
16
)
= 9 ±
9
−
8
x
−
x
2
.
In other coordinate systems, for example, polar coordinates, the equation depends on other variables, related to x and y in some way, so the equation may look quite different.
To see results in polar form we need to substitute expressions in the new variables in place of the original x and y. So, = ⋅cos ; = ⋅sin
x
=
ρ
⋅
cos
θ
;
y
=
ρ
⋅
sin
θ
, and equation (1) becomes(⋅cos + 4)2 + (⋅sin − 9)2 = 25 (ρ⋅cos
θ
+ 4
)
2
+ (ρ⋅sin
θ
− 9
)
2
= 25
This quickly becomes messy when we try solving for one of the new variables. I won’t bother you with too much trig., but, one possible version of our equation becomes: = 9sin−4cos ± 9 − 72cossin − 47cos2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√
ρ
=
9
sin
θ
−
4
cos
θ
±
9
−
72
cos
θ
sin
θ
−
47
cos
2
θ
Now, if we encountered these equations in the wild, we may clearly see some similarities. They all involve squares and square-roots. The numbers nine and four float around through them, etc. But, would we see that they define the same geometric object?
