what is the equation of circle whose center is (-3,-2) and passes through points(2,1)

The genera formula for circle equation is (x-a)^2+(x-b)^2=r^2
such that (a,b) center of the circle and r is the radius. In this question there is no radius given.but we have the center and one point passes. Then the equation will be
(x-(-3))^2+(y-(-2))^2=r^2... (1)
now take the point (2,1), substitution x=2 and y=1 in(1), we have
(5)^2+((3)^2=r^2
25+9=r^2
25=r^2
r=5, we can't take another value for r because it is in negative and we deal with distance, then r=5.
substitution r=5 in (1), we have
(x-(-3))^2+(y-(-2))^2=5^2
(x+3)^2+(y+2)^2=25
Best wishes
Mr.Ahmad Rababah
Najran University

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