Question

ss below

Answer 1

circle is the locus of apoint which moves such that its locus from a fixed point always remains constant.
if (h,k) is the fixed point and let (x,y) be any point on the locus,then
\[\left| \sqrt{x-h)^2+(y-k)^2}=r \right|\]\[where~r~is~the~fixed~distance \]
squaring
\[(x-h)^2+(y-k)^2=r^2\]
fixed point is called the centerand fixed distance=radius.

Answer 2

im still confused

Answer 3

first find the distance from the center to a point on the circle
here (x,y)=(1,8),(h,k)=(-3,5)
so find r

Answer 4

how do i find r

Answer 5

\[r=\sqrt{(1-(-3))^2+(8-5)^2}=\sqrt{(1+3)^2+3^2}=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5\]
now write the eq. of circle.

Answer 6

can you write or still have aproblem.

Answer 7

\[(x-(-3)^2+(y-5)^2=5^2\]
or
\[(x+3)^2+(y-5)^2=25\]

Answer 8

i got it thx

Answer 9

yw