Find the equation of the circle whose centre is the point (3,-2) and which passes through the?

This is just busy work, something to occupy your attention for a while. First find the center of the given circle.

4x^2 + 4y^2 + 8x + 2y + 1 = 0 Add spaces to improve readability.

The equation of a circle is x^2 + y^2 = r^2 and that is centered on the origin. If you want it centered on some point (a, b) you subtract the coordinates: (x - a)^2 + (y - b)^2 = r^2. So your first step is to resolve the given equation into that form.

A big part of math is pattern recognition. A lot of homework is just fighting with stuff so you will remember the pattern when you see it again. The most common pattern is (x + a) * (x + b) = x^2 + (a + b)x + ab and the special case (x + a) * (x - a) = x^2 - a^2. When you spot the pattern you can just write the answer from memory.

4x^2 + 8x + 4 = 4(x + 1)^2
4y^2 + 2y + 1 = 4(y + 1/4)^2

So: 4(x + 1)^2 + 4(y + 1)^2 = -1 + 5 We had to subtract 1 from both sides to get the form we wanted, and then add 5 because we added 5 on the left side to complete the squares. That is not important because we only wanted the center point and that is (-1, -1/4). Now you write the equation of a circle whose center is (3, -2) and radius is the distance from there to (-1, -1/4).

r^2 = (3 + 1)^2 + (-2 + 1/4)^2 = 19.0625

(x + 1)^2 + (y + 1/4)^2 = 19.0625 <-- ANSWER

Check: www.wolframalpha.com/input/?i=plot+4x%5E2+%2B+4y%5E2+%2B+8x+%2B+2y+%2B+1+%3D+0