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Find the center, radius, and x- and y-intercepts of the circle given by the equation 3x2 + 3y2 - 6x + 5y = 0.
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Divide thru by 3 to get:
x^2-2x+y^2+(5/3)y=0
Complete the square, as follows:
(x^2-2x+1)+(y^2+(5/3)y+(5/6)^2)=1+(25/36)
(x-1)^2+(y+(5/6))^2 =61/36
Center: (1,-(5/6))
Radius: [sqrt(61)]/6
x-intercepts:
Let y=0 then x^2-2x=0; then x=0 or x=2; Intercepts at (0,0) and (2,0)
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y-intercepts:
Let x=0, then y^2+(5/3)y=0; y=0 or y=(-5/3): Interceps at (0,0), and (0,-5/3)
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3(x^2-2x+1)+3(y^2+(5/3)y+(5/6)^2) = 3+3(5/6)^2
3(x-1)^2 + 3(y+(5/6))^2 = 108/36 + 25/36
3(x-1)^2 + 3(y+(5/6))^2 = 133/36
(x-1)^2 + (y+(5/6))^2= 133/108



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GIVE THE CENTER OF THE CIRCLE WITH EQUATION x^2+2x+y^2-10y+22=0.
The standard form of the equation of a circle is {{{highlight((x-h)^2+(y-k)^2=r^2)}}}, (h,k)=center, r=radius
We complete the squares in order to put the equation in standard form.
x^2+2x+y^2-10y+22-22=0-22
x^2+2x+____+y^2-10y+_____=-22+_____+______
To complete the squares take (1/2) the coefficient of the middle term and add it to both sides of the equation.
x^2+2x+(2/2)^2+y^2-10y+(-10/2)^2=-22+(2/2)^2+(-10/2)^2
x^2+2x+(1)^2+y^2-10y+(-5)^2=-22+(1)^2+(-5)^2
x^2+2x+1+y^2-10y+25=-22+1+25
(x+1)^2+(y-5)^2=4
The center (h,k)=(-1,5)
If you were asked for the radius, it would be {{{sqrt(4)=2}}}
Happy Calculating!!!


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Find the standard equation of the circle that satisfies the conditions.

Center (-1,-3), passing through the point (3,0)

In order to write an equation for a circle you need a center and a radius.  You're missing a radius, but you can find it with the information that they gave you.  The radius is the distance from the center to any point of the circle.
The distance formula is r={{{highlight(d=sqrt((x2-x1)^2+(y2-y1)^2))}}}
(x1,y1)=(-1,3) and (x2,y2)=(3,0)
r={{{d=sqrt((3-(-1))^2+(0-3)^2)}}}
r={{{d=sqrt((4)^2+(-3)^2)}}}
r={{{d=sqrt(16+9)}}}
r={{{d=sqrt(25)}}}
r={{{d=5}}}
Now that we have a center (h,k)=(-1,-3) and a radius r=5.
The standard form of a circle:  {{{(x-h)^2+(y-k)^2=r^2}}}, (h,k) is the center, and r is the radius.
{{{(x-(-1))^2+(y-(-3))^2=5^2}}}
{{{highlight((x+1)^2+(y+3)^2=25)}}}
:
Happy Calculating!!!


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Find the centre and radius of the circle:
{{{x^2 + y^2 - 12x - 16y - 21 = 0}}} First, add 21 to both sides.
{{{x^2 + y^2 - 12x - 16y = 21}}} Now group the x-terms together and the y-terms together.
{{{(x^2 - 12x) + (y^2 - 16y) = 21}}} Next, complete the square in x by adding the square of half the x-coefficient to both sides of the equation. {{{(-12/2)^2 = 36}}}, and the same for the y {{{(-16/2)^2 = 64}}}
{{{(x^2 - 12x + 36) + (y^2 - 16y + 64) = 21 + 36 + 64}}} Simplify.
{{{(x^2 - 12x + 36) + (y^2 - 16y + 64) = 121}}} Factor the x-group and the y-group.
{{{(x - 6)^2 + (y - 8)^2 = 121}}}  Take the square root of 121 = 11
{{{(x - 6)^2 + (y - 8)^2 = (11)^2}}} Compare with the standard form for a circle with centre at (h, k) and radius, r.
{{{(x - h)^2 + (y - k)^2 = r^2}}}

The circle centre is at (6, 8) and the radius is 11

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Given the circle: (y+7)^2+(x-4)^2=18
Find the center: (4,-7)
Find the exact radius: {{{3sqrt(2)}}}
:
The standard form of the equation of a line is {{{highlight((x-h)^2+(y-k)^2=r^2)}}}, Where the center is (h,k) and the radius is r.
Your equation is a little out of order and I initially put down the wrong answer for the center because I didn't notice.
(x-4)^2+(y+7)^2=18
The center is:
{{{(x-highlight(4))^2+(y-highlight(-7))^2=18}}} (h,k)=(4,-7)
The radius is:
{{{(x-4)^2+(y+7)^2=highlight(18)}}}  
{{{r^2=18}}}
{{{r=sqrt(18)}}}
{{{r=sqrt(9)*sqrt(2)}}}
{{{r=3*sqrt(2)}}}
Happy Calculating!!!



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Remember that the area of a circle is given by:
{{{A = (pi)r^2}}} To solve for r (radius), divide both sides by {{{(pi)}}}
{{{A/(pi) = r^2}}} Now take the square root of both sides.
{{{sqrt(A/(pi)) = r}}}

If A = 25 sq. meters, and {{{(pi)}}} = 3.14 (Approximately)
{{{r = sqrt(25/3.14)}}}
{{{r = 2.821}}}meters, rounded to the nearest thousandth.

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