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(x+3)^2+(y-4)^2=16please provide the center and the radius

Sagot :

Given:

Given the equation of the circle

[tex](x+3)^2+(y-4)^2=16[/tex]

Required: Radius and center of the circle

Explanation:

The standard form of an equation of a circle is of the form

[tex](x-h)^2+(y-k)^2=r^2[/tex]

where (h, k) is the center and r is the radius.

Re-write the given equation of circle in standard form.

[tex](x-(-3))^2+(y-4)^2=4^2[/tex]

Comparing with the standard form,

center: (h, k) = (-3, 4)

Radius: r = 4

Final Answer: Center = (-3, 4) and radius = 4.

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