GIVING 100 POINTS ...Pls provide an expatiation on what I did wrong. Thank you

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?

Wow. that is a really cool question.