Answer:
The circle equation that has center , (1,-3) and passes through , (-6,-2) is given as [tex](x-1)^2 + (y+3)^2 = 50[/tex]
Step-by-step explanation:
Here, the coordinate (h,k) of the center of circle = (1,-3)
Also, the point (x,y) on the circle = (-6,-2)
Let us assume the radius of the circle = r
The General equation of Circle is : [tex](x-h)^2 + (y-k)^2 = (r)^2[/tex]
Substituting the value of (h,k) and (x,y) , we get:
[tex](-6-1)^2 + (-2-(-3))^2 = r^2\\\implies (-7)^2 + (-1)^2 = r^2\\\implies 49 + 1 = r^2\\\implies r= \sqrt{50} = 7.07[/tex]
So, the radius of the circle = 7.07 units
Now, substitute the value of (h,k) = (1,-3) and r = 7.07 back in to the general circle equation, we get:
[tex](x-1)^2 + (y-(-3))^2 = (7.07)^2[/tex]
[tex]\implies (x-1)^2 + (y+3)^2 = 50[/tex]
Hence, the circle equation that has center , (1,-3) and passes through , (-6,-2) is given as [tex] (x-1)^2 + (y+3)^2 = 50[/tex]
