find (h,k) and r of the circle (2x^2)-4x+(2y^2)-10y-10x=0

Question

conic systems circles

Answer 1

It seems like there is a typo in the problem.  

When I correct it, it would be

 

2x² -4x +2y² -10y -10 = 0

 

Then, divide both sides by 2

 

x² -2x +y² -5y -5 = 0

 

(x-1)²-1 + (y-5/2)² -25/4 -10 =0

 

(x-1)²+ (y-5/2)²  = 69/4

 

So, the center is located at (1,5/2) and the radius is sqrt(69)/2

Answer 2

I'm assuming you mean 2x^2-4x+2y^2-10y-10=0, not 10x

divide all terms by 2

x^2-2x+y^2-5y-5=0

x^2-2x+y^2-5y=5

the equation of a circle is (x-a)^2+(y-b)^2=c^2

the center is (a,b) and the radius is c

(x^2-2x+a)+(y^2-5y+b)=5+a+b

(2/2)=1 and (5/2)=2.5

1^2=1 and (2.5)^2=6.25

let a=1 and b=6.25

we have (x^2-2x+1)+(y^2-5y+6.25)=5+1+6.25

(x-1)^2+(y-2.5)^2=12.25

(x-1)^2+(y-2.5)^2=(3.5)^2

the center is (1,2.5) and the radius is 3.5