# Math Help - Cyclic quad

Circle radius 65 has center M(65,65), thus tangent to both x and y positive axis.

ABCD is a cyclic quad, it's vertices on the circumference of circle M.

It's area is 6072 square units.

Coordinates of A are (2,81), coordinates of C are (98,9).

What are the coordinates of B and D?

In case some of you suspect I'm "fishing" for an answer:
the x and y coordinates of B and D add up to 260 (that's from my solution).

I'm posting this for your pleasure(!), hoping nobody finds more than 1 solution

Are the coordinates positive integers or any real numbers is acceptable?

Positive integers.

A(2,81)
B(2,49)
C(98,9)
D(128,81)
But as you can switch B and D there are two answers to this question.

Thanks; mine was:
A(2,81), B(98,121), C(98,9), D(32,9) ..... clockwise !

Distance between A and B: $a=104$
Distance between B and C: $b=112$
Distance between C and D: $c=66$
Distance between D and E: $d=78$
Using Brahmagupta's formula:
$A=\sqrt{(s-a)(s-b)(s-c)(s-d)}$
Where semiperimeter $s$ is defined as $s=\frac{a+b+c+d}{2}$
So with your numbers, semiperimeter s would be:
$s=\frac{104+112+66+78}{2}=180$
And Area would be:
$A=\sqrt{(180-104)(180-112)(180-66)(180-78)}=7752$
Am I doing something wrong or did you miss something somewhere?