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1. The equation of a circle is x^2+y^2+10x-4y-20=0.

What is the radius of the circle?

 

2. The general form of the equation of a circle is x^2+y^2+2x-6y+1=0.

What are the coordinates of the center of the circle?

 

3. The standard form of the equation of a circle is (x-4)^2+(y-2)^2=9.

What is the general form of the equation?

 

Options: x^2+y^2+8x+4y-29=0,  x^2+y^2-8x-4y+11=0,  x^2+y^2+8x+4y+11=0,  or x^2+y^2-8x-4y-29=0

 

Thanks in advance! Also, I just need help, right now I currently have a solid 'B' in my geometry/trigonometry class so rude comments are not needed. Thanks

Guest Apr 26, 2017
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1. Add 25 to both sides of the equation and collect as follows:

 

x^2 + 10x + 25 + y^2 - 4y - 20 = 25.  This can be written as:

(x + 5)^2 + y^2 - 4y - 20 = 25

 

Now add 4 to both sides and collect as follows:

(x + 5)^2 + y ^2 - 4y + 4 - 20 = 29. This can be written as:

(x + 5)^2 + (y - 2)^2 - 20 = 29

 

Add 20 to both sides:

(x + 5)^2 + (y - 2)^2 = 49

 

Now the standard form is (x - xc)^2 + (y - yc)^2 = r^2 where (xc, yc) are the coordinates of the centre and r is the radius.

 

So we have a circle with centre (-5, 2) and radius 7 (because 7^2 = 49)

 

2. You can now use the same approach here. To complete the square for the x values add the square of half the coefficient of the x term to both sides (similarly for completing the square of the y terms).

Alan  Apr 26, 2017

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