lingkaran x²+y²+6x+8y+p=0 melalui titik (1,1) berapa jari² nya

pada titik (1,1) -> 1² + 1² + 6.1 + 8.1 +p = 0
                        -> 1 + 1 + 6 + 8 + p =0
                        -> 16 +p = 0
                        -> p = -16
pers: x² + y² + 6x + 8y -16 = 0

A= 6, B = 8, C = -16

R =[tex] \sqrt{ \frac{1}{4} A^{2} + \frac{1}{4} B^{2} - C} [/tex]
R =[tex] \sqrt{ \frac{1}{4} 6^{2}+ \frac{1}{4} 8^{2}-(-16) } [/tex]
R = √(9+16+16) = √41