Question

VERY URGENT HAVE A TEST TO TAKE

What is the general form of the equation of a circle with its center at (-2,1) and passing through (-4,1)?

a) x^2+y^2-4x+2y+1=0

b) x^2+y^2+4x-2y+1=0

c) x^2+y^2+4x-2y+9=0

d) x^2-y^2+2y+y+1=0

Answer 1

the easiest way is to try the point (-4,1), that is, x=-4, y=1,
to see which equation works.
b works.

The usual way to do it is to find the equation of the circle
standard form of a circle is (x-h)²+(y-k)²=r², (h,k) are the coordinates of the center, r is the radius.
in this case, the center is (-2,1), so (x+2)²+(y-1)²=r²
the given point (-4,1) is for you to find r: (-4+2)²+(1-1)²=r², r=2
so the equation is (x+2)²+(y-1)²=2²
expand it: x²+4x+4+y²-2y+1=4
x²+y²+4x-2y+1=0, which is answer b.