1) Find the center and radius of the circle given by the equation x^2+y^2-4x-6y+

Question

1) Find the center and radius of the circle given by the equation
x^2+y^2-4x-6y+12=0

he center is : (2,3)
The radius is : ______
write in standard form (x-h)^2-(y-k)^2=r^2
The same equation in the form is
(x-2)^2+(y-3)^2 =_________


2)Find the center and radius of the circle given by the equation
x^2+y^2+4x+2y-4=0

The center is : (-2,-1)
The radius is : ______


equation in the form (x-h)^2+(y-k)^2=r^2
(x+2)^2 + (y+1)^2 =_____

3)

(1 pt) Find the standard form for the equation of a circle (x-h)^2+(y-k)^2=r^2 with a diameter that has endpoints and .

the equation is _____+_____=_____

Explanation / Answer

Find the standard form for the equation of a circle (x-h)^2+(y-k)^2=r^2 with a diameter that has endpoints (-1,-7) and (6,6) the equation is _____+_____=_____ Center of the circle: (x1+x2)/2, (y1+y2)/2 (-1+6)/2 = h, (-7+6)/2 = k Solve and plug these into the equation above To find r you need to find the distance d=sqrt[(x2-x1)^2+(y2-y1)^2] d=sqrt[(6-(-1))^2 + (6-(-7)^2] d=sqrt[49+169] = 14.76 divide d by two and you get your answer. Please do not copy and really spend time looking at this and try to understand it. I spent well over 30 minutes working on these problems--for 20 karma points many people would have overlooked your request. However, I like it when people benefit from my help so PLEASE try to understand this material and refer back to this when you get a problem like this again. cheers

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