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1)The equation of a circle is x2+y2−12x+6y+20=0 .

What is the radius of the circle?

Enter your answer in the box.

r =___  units

 

2)

The general form of the equation of a circle is x2+y2+2x−6y+1=0.

 

What are the coordinates of the center of the circle?

 

Enter your answer in the boxes.

(___,___)

 

3)

The standard form of the equation of a circle is (x−3)2+(y−1)2=16.

 

What is the general form of the equation?

x2+y2−6x−2y−6=0

x2+y2+6x+2y+26=0

x2+y2−6x−2y−26=0

x2+y2+6x+2y−6=0

 

4)

The center of a circle is located at (6, −1) . The radius of the circle is 4.

 

What is the equation of the circle in general form?

x2+y2+12x−2y+33=0

x2+y2−12x+2y+33=0

x2+y2+12x−2y+21=0

x2+y2−12x+2y+21=0

 

5) see image

What is the equation of this circle in general form?

x2+y2−6x+4=0

x2+y2−6x−16=0

x2+y2+6x−16=0

x² + y² + 6x + 4 = 0

CrazyDaizy  May 17, 2017
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can anyone help answers these?

CrazyDaizy  May 19, 2017
 #2
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1) \(r=5units\)

 

Let's see why:

 

First rearrange the terms to make it easier to work with:

\(x^2-12x+y^2+6y=-20\)

 

I have grouped the x- and y-containing terms together. Now, we must use a method called completing the square. In a quadratic term, the equation is ax^2+bx+c=0. In the above equation, however, there is no c-term. To find the missing c-term, do (b/2)^2.

 

\(x^2-12x+(\frac{b_1}{2})^2+y^2+6y+(\frac{b_2}{2})^2=-20+(\frac{b_1}{2})^2+(\frac{b_2}{2})^2\)  Remember, whatever you do to one side, you must do to the other.

\(x^2-12x+(\frac{-12}{2})^2+y^2+6y+(\frac{6}{2})^2=-20+(\frac{-12}{2})^2+(\frac{6}{2})^2\) Substituting the values 

\(x^2-12x+36+y^2+6y+9=-20+36+9\)

\(x^2-12x+36+y^2+6y+9=25\)

 

Okay, we have manipulating the equation so that the number on the right side of the equation is the r^2, You need not go further with the simplification of this problem as it is only asking for the radius length. We have enough information for this

 

\(r^2=25\)

\(r=5units\)Of course, include units in the final answer.

 

2) \(C=(1,3).\)

 

Here's why:

 

Just like the previous problem, I will rearrange the terms, so the equation ios easier to handle:

\(x^2-2x+y^2-6y=-1\)

 

We will use the same process as the previous problem: 

 

\(x^2-2x+(\frac{b_1}{2})^2+y^2-6y+(\frac{b_1}{2})^2=-1+(\frac{b_1}{2})^2+(\frac{b_2}{2})^2\)

\(x^2-2x+(\frac{-2}{2})^2+y^2-6y+(\frac{-6}{2})^2=-1+(\frac{-2}{2})^2+(\frac{-6}{2})^2\)

\(x^2-2x+1+y^2-6y+9=-1+1+9\)

\(x^2-2x+1+y^2-6y+9=9\)

 

In this case, the problem is asking for the center of the circle. The standard form for a circle is (x – h)2 + (y – k)2 = r2

, where (h,k) is the center of the circle. However, we still need to manipulate the equation further. :Luckily, x^2+2x+1 and y^2-6y+9 are both perfect square trinomials, which means that they can be represented as (x+(b/2))^2 and (y+(b/2))^2.

 

\((x+\frac{b_1}{2})^2+(y+\frac{b_2}{2})^2=9\)The number that goes for b is always the b coefficient in quadratics. 

\((x+\frac{-2}{2})^2+(y+\frac{-6}{2})^2=9\)Substituting

\((x-1)^2+(y-3)^2=9\)Finally, after all this work, we have gotten to the standard equation of a circle. 

 

\(-1=-h\),so \(h=1\).

\(-3=-k\),so \(k=3\).

 

The center of a cricle is always \((h,k)\),so the center, or C, is (1,3).

Guest May 19, 2017

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