what is the standard form of x2+y2-4y+3=0

Question

getting the coordinates of the center and the radius

Answer 1

i recognise this as the equation of a circle, and this is also hinted at in the question because of the words "centre" and "radius".

The coefficients of x and y are the same, being 1, which confirms a circle rather than an ellipse.

What we have to do is to complete the square for y because we have a y term but no x term.

So: x^2 + (y^2-4y+4) -4+3=0; x^2+(y-2)^2-1=0; x^2+(y-2)^2=1 is the standard form.

On the right we have the radius squared so the radius is 1.

The centre is (0,2), because, when x=0 y is 3 or 1, and when y=2, x=1 or -1. The centre lies in between these extremes: halfway between 1 and -1 is x=0, and halfway between 3 and 1 is y=2.