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

... centre of the circle 4x^2+4y^2+8x+2y+1=0 or:... centre of the circle 4x^2+4y^2+8x+2y+1=0or:This is just busy work, something to occupy your atte

... centre of the circle 4x^2+4y^2+8x+2y+1=0

or:... centre of the circle 4x^2+4y^2+8x+2y+1=0

or: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)^24y^2 + 2y + 1 = 4(y + 1/4)^2So: 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

or:Simplifying4x2 + y2 + -8x + -2y + 1 = 0Reorder the terms:1 + -8x + 4x2 + -2y + y2 = 0Solving1 + -8x + 4x2 + -2y + y2 = 0Solving for variable 'x'.The solution to this equation could not be determined.

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