Solution 4.1:7c

From Förberedande kurs i matematik 1

By completing the square, we can rewrite the x- and y-terms as quadratic expressions,

x2−2xy2+6y=(x−1)2−12=(y+3)2−32

and the whole equation then has standard form,

(x−1)2−1+(y+3)2−9(x−1)2+(y+3)2=−3=7.

From this, we see that the circle has its centre at (1,-3) and radius 7 .