Question

In the following find the equation of the circle with centre (1, 1) and radious $\sqrt{2}$.

Answer 1

here is answer take the following step( s) into consideration !

we know that general equation of circle is given by => x² + y² + 2gx +2fy + c


c = √2 -√2 = 0

point = (1,1 )

so , we have :-

x² + y² + 2 g + 2f + 0

or x² + y² + 2g + 2f

hope it help you !!

thanks !!

Answer 2

Answer:

Equation of the circle:
 $x^{2} +y^{2} -2x-2y=0$

Step-by-step explanation:

Standard equation of a circle:$(x-a)^{2} +(y-b)^{2} =r^{2}$

here a and b are the centre of circle,and √2 is the radius of the circle.

So put a =1

b=1

r=√2

$(x-1)^{2}+(y-1)^{2}  =(\sqrt{2})^{2} \\ \\ x^{2} -2x+1+y^{2} -2y+1=2\\ \\ x^{2} +y^{2} -2x-2y=0$