+99 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
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).
