Answer:
A
Step-by-step explanation:
if (x1,y1) and (x2,y2) are the extremities of diameter,then eq. of circle is
(x-x1)(x-x2)+(y-y1)(y-y2)=0
reqd. eq. is (x+1)(x-5)+(y+9)(y-1)=0
[tex]x^{2} +x-5x-5+y^2+9y-y-9=0\\x^{2} -4x+y^2+8y-14=0\\compare with x^2+y^2+2gx+2fy+c=0\\center (-g,-f) \\radius r=\sqrt{g^2+f^2-c}[/tex]
center is (2,-4)
r=√(2²+(-4)²-(-14))
=√(4+16+14)
=√(34)
eq. of circle is (x-2)²+(y+4)²=34
or
(x²-4x)+(y²+8y)=14
(x²-4x+4)+(y²+8y+16)=14+4+16
(x-2)²+(y+4)²=34