Respuesta :
Since, the equation involved is a circle let us clearly write the equation first:
[tex](x+5)^2+(y+2)^2=21^2[/tex]
From the standard equation of the circle:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
[tex](h,k)=(x,y)[/tex]
So from the equation given above, our x would be -5, and our y is equivalent to -2.
So the center of our circle is located at the point (-5,-2)
[tex](x+5)^2+(y+2)^2=21^2[/tex]
From the standard equation of the circle:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
[tex](h,k)=(x,y)[/tex]
So from the equation given above, our x would be -5, and our y is equivalent to -2.
So the center of our circle is located at the point (-5,-2)
Answer:
(-5,-2)
Step-by-step explanation:
the equation involved is a circle let us clearly write the equation first:
From the standard equation of the circle:
So from the equation given above, our x would be -5, and our y is equivalent to -2.
So the center of our circle is located at the point (-5,-2)