How do you write an equation of a circle with center (-2,1) and radius with endpoint at (1,0)?

The simplest equation for a circle is:
#color(white)("XXX")X^2+Y^2=R^2#
for a circle with center at #(0,0)# and radius #R#

If we want to shift this so the center is at #(-2,1)#
then the #X# values become #x=X-2color(white)("xx")rarrcolor(white)("xx")X=x+2#
and
the #Y# values become #y=Y-1color(white)("xx")rarrcolor(white)("xx")Y=y-1#
So #X^2+Y^2# becomes #(x+2)^2+(y-1)^2#

The radius is une3ffected by the shift and will remain the same length.
The radius is the distance between the center #(-2,1)# and any point on the circumference; in this case we are given the point #(1,0)#
Using the Pythagorean Theorem this gives us a radius squared of
#color(white)("XXX")R^2=((-2)-1)^2+(1-0)^2=10#

Therefore the equation of our (shifted) circle will be
#color(white)("XXX")(x+2)^2+(y-1)^2=10#

#"the standard form of the equation of a circle is "#

#color(red)(bar(ul(|color(white)(2/2)color(black)((x-a)^2+(y-b)^2=r^2)color(white)(2/2)|)))#

#"where "(a,b)" are the coordinates of the centre and r"#
#"is the radius"#

#"the distance from the centre to the endpoint gives r"#

#"to calculate r use the "color(blue)"distance formula"#

#•color(white)(x)r=sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

#rArrr=sqrt((-2-1)^2+(1-0)^2)=sqrt10#

#(x-(-2))^2+(y-1)^2=(sqrt10)^2#

#rArr(x+2)^2+(y-1)^2=10larrcolor(red)"equation of circle"#