Question

Find the centre and radius of the equation of the circle. 2x^2+2y^2-8x+5y+10=0

Answer 1

Solution. Transform the equations of the circle

$$2x^2+2y^2-8x+5y+10=0$$ $$2(x-2)^2-8+2(y+\frac{5}{4})^2-\frac{25}{8}+10=0$$ $$2(x-2)^2+2(y+\frac{5}{4})^2=\frac{9}{8}$$ $$(x-2)^2+(y+\frac{5}{4})^2=\frac{9}{16}$$

General equation of a circle

$$(x-a)^2+(y-b)^2=R^2$$

where (a,b) is circle center coordinates; R is circle radius.

Therefore circle center coordinates

$$(2,-\frac{5}{4})$$

circle radius

$$R=\frac{3}{4}$$

Answer.

$$(2,-\frac{5}{4})$$ $$R=\frac{3}{4}$$