homework help 1692

##0 = x^2+y^2-8x+6y+16 = (x-4)^2 + (y+3)^2 – 3^2##

is a circle of radius ##3## with centre ##(4, -3)##

The equation of a circle of radius ##r## centred at ##(a, b)## can be written:

##(x-a)^2+(y-b)^2 = r^2##

We are given:

##0 = x^2+y^2-8x+6y+16##

##=x^2-8x+16 + y^2+6y+9 – 9##

##=(x-4)^2 + (y+3)^2 – 3^2##

So:

##(x-4)^2+(y+3)^2 = 3^2##

which is in the form of the equation of a circle of radius ##3## centre ##(4, -3)##

graph{(x-4)^2+(y+3)^2 = 3^2 [-7.875, 12.125, -7.8, 2.2]}