SOLUTION: Find the equation of each circle described.
Passes through the origin and is concentric with the circle x^2-6x+y^2-4y+4=0
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Question 214271: Find the equation of each circle described.
Passes through the origin and is concentric with the circle x^2-6x+y^2-4y+4=0
Found 2 solutions by Alan3354, solver91311:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Passes through the origin and is concentric with the circle x^2-6x+y^2-4y+4=0
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Find the center of this circle
x^2-6x+y^2-4y+4=0
x^2-6x+9 + y^2-4y+4 = 9
(x-3)^2 + (y-2)^2 = 3^2
That's a circle of radius 3 with a center at (3,2)
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The other circle has the same center (concentric). Find its radius:
r^2 = 3^2 + 2^2
r^2 = 13
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--> (x-3)^2 + (y-2)^2 = 13
or, x^2-6x+y^2-4y=0
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The equation of a circle centered at )
with radius 
is:
^2 + (y - k)^2 = r^2)
Let's begin by determining the center of the given circle, which is the center of the circle we want because they are concentric.
Take the equation of the given concentric circle:
and complete both squares. First put the constant term on the right.
Remember for a squared binomial, the third term of the resulting trinomial is one-half of the second term coefficient squared. -6 divided by 2 is -3; -3 squared is 9, so add 9 to both sides. And -4 divided by 2 is -2; -2 squared is 4, so add 4 to both sides.
Now factor the two trinomials:
^2 + (y - 2)^2 = 9)
Hence, the the given circle is centered at (3, 2) with radius 3.
Since the desired circle passes through the origin, the line segment with endpoints at the origin and at (3, 2) is a radius of the desired circle. We need to know the measure of the radius which is the distance from (0, 0) to (3, 2). Use the distance formula:
^2 + (2 - 0)^2} = sqrt{9 + 4} = sqrt{13})
Now that we know the center and radius of the desired circle we can write the equation:
^2 + (y - 2)^2 = \sqrt{13}^2 = 13)
Expand the binomials:
Simplify:
John
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