(1)
x2+y2−10x−12y+6=0
Now by completing the squares
x2−10x+(210)2+y2−12y+(212)2+6=(210)2+(212)2⟹(x2−10x+25)+(y2−12y+36)=2+36−6⟹(x−5)2+(y−6)2=55..........................(1)
standard form of the circle is
(x−h)2+(y−k)2=r2...................................(2)
where (h,k)is the center and r is radius of the circle by comparing ed. (1) and (2)
we get,
centre(h,k)=(5,6)
Radiusr=55
(2)
−2x2−2y2−18y+9=0⟹−2(x2+y2+9y−29)=0⟹x2+y2+9y−29=0
by completing squares,
x2+y2+9y+(29)2=29+(29)2⟹x2+y2+9y+481=29+481⟹x2+(y+29)2=418+81⟹(x−0)2+(y+(−29))2=499
By comparing from standard form of circle we get,
centre (h,k) =(0,2−9)
Radius r=499=4311
Comments