Well, honey, the center of that circle is simply the point (3, 9). You see, the equation you provided is in the form (x - h)² + (y - k)² = r², where (h, k) is the center of the circle. So, in this case, the center is at (3, 9). That's all there is to it, sugar.
If you mean: (x-3)^2 +(y-9)^2 = 16 then the center of the circle is at (3, 9)
(-7,5)
x2 + y2 = 16
The equation provided appears to have a typographical error, as it should likely be in the form of a standard circle equation. If you meant (x^2 + y^2 = 16), the center of the circle is at the coordinates (0, 0). If this is not the correct interpretation, please clarify the equation for an accurate response.
(x+7)2+(y+5)2 = 16 Centre of circle: (-7, -5) Radius: 4
The center of the circle is at (-7, 5)
If you mean: (x-3)^2 +(y-9)^2 = 16 then the center of the circle is at (3, 9)
(-7,5)
x2 + y2 = 16
The equation provided appears to have a typographical error, as it should likely be in the form of a standard circle equation. If you meant (x^2 + y^2 = 16), the center of the circle is at the coordinates (0, 0). If this is not the correct interpretation, please clarify the equation for an accurate response.
It's a circle. The equation of a circle is x^2+y^2=r^2. So the equation you've given is a circle with a radius of 4 and, since there are no modifications to the x or y values, the center of the circle is located at (0,0).
(x+7)2+(y+5)2 = 16 Centre of circle: (-7, -5) Radius: 4
center 5,-3 radius 4
To find the center and radius of the circle given by the equation (x^2 + y^2 - 8x - 4y - 16 = 0), we first rewrite it in standard form. Completing the square for both (x) and (y) gives us ((x - 4)^2 + (y - 2)^2 = 36). Thus, the center of the circle is at ((4, 2)) and the radius is (6) (since (r = \sqrt{36})).
(x-0)² + (y-0)² = r²
It is: (x+3)^2 + (y-5)^2 = 16
(x - 9)2 + (y + 5)2 = 16