Question

Which is an equation of a circle with center (5, 0) that passes through the point (1, 1)

Answer 1

Equation of circle at center (h,k) is given by

$(x-h)^2+(y-k)^2=r^2$

Given that center is at (5,0) that means h=5 and k=0

plug both values into above formula

$(x-5)^2+(y-0)^2=r^2$

$(x-5)^2+y^2=r^2$ ...(i)

Given that circle passes through point (1,1) so it will satisfy above equation

$(1-5)^2+1^2=r^2$

$(-4)^2+1^2=r^2$

$16+1=r^2$

$17=r^2$

Now plug this value of r^2 into equation (i)

$(x-5)^2+y^2=17$

which best matches with choice C

Hence $(x-5)^2+y^2=17$ is the final answer.