what is the equation of the circle passing through the point (6,5) and centered at (3,-4)?

1. (x-6)²+(y-5)²=82
2. (x-6)²+(y-5)²=90
3. (x-3)²+(y+4)²=82
4. (x-3)²+(y-4)²=90

The equation of a circle with center (h,k) and radius r is
(x-h)²+(y-k)²=r²
so
given center is (3,-4)
(x-3)²+(y-(-4))²=r²
(x-3)²+(y+4)²=r²
find r² by subsitutiong the point taht it passes through which is (6,5)

so
(6-3)²+(5+4)²=r²
3²+9²=r²
9+81=r²
90=r²
so

(x-3)²+(y+4)²=90 is da equation

4th equation