Question

The point (1,-3) is on a circle with center (-2,1). Write the equation of the circle and graph it as ⊙ C.

Answer 1

the eqn of a circle, center (a,b):
\[(x-a)^{2}+(y-b)^{2}=r ^{2}\]

Answer 2

So then (x+2)2+(y-1)=r2?

Answer 3

The center of the circle here is (-2,1),
so a= -2 and b=1

therefore, plugging in the values
\[(x-(-2))^{2}+(y-1)^{2}=r ^{2}\]
\[(x+2)^{2}+(y-1)^{2}=r ^{2}\]

now you need to figure out what r sqrd is

Answer 4

Would I use the other coordinates for that?

Answer 5

yeah so draw a triangle

and solve using Pythagoras

Answer 6

\[r ^{2}=3^{2}+4^{2}\]
\[r ^{2}=25\]

\[(x+2)^{2}+(y-1)^{2}=r ^{2}\]
replace r sqrd with 25
\[(x+2)^{2}+(y-1)^{2}=25\]
and that is yr final ans