Answer:

Step-by-step explanation:
The standard equation for circle is

where point (a,b) is coordinate of center of circle and r is the radius.
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Given
center of circle =((-2,3)
let r be the radius of circle
Plugging in this value of center in standard equation for circle given above we have

Given that point (1,2 ) passes through circle. Hence this point will satisfy the above equation of circle.
Plugging in the point (1,2 ) in equation 1 we have

now we have value of r^2 = 10, substituting this in equation 1 we have
Thus complete equation of circle is 