Write an equation for a circle centered at the point (–9, 3) with radius 8.
Why did the circle go to a center point (–9, 3) with a radius of 8? Because it wanted to be perfectly balanced and not go off on a tangent! Okay, here's the equation for you:
(x + 9)² + (y - 3)² = 64
To write the equation for a circle centered at the point (–9, 3) with radius 8, we will use the standard form equation of a circle. The equation is given as:
(x - h)^2 + (y - k)^2 = r^2
Where (h, k) represents the coordinates of the center of the circle, and r represents the radius of the circle.
In this case, the center of the circle is (–9, 3) and the radius is 8. Plugging these values into the equation gives us:
(x - (-9))^2 + (y - 3)^2 = 8^2
Simplifying further:
(x + 9)^2 + (y - 3)^2 = 64
Thus, the equation for the circle centered at (–9, 3) with radius 8 is (x + 9)^2 + (y - 3)^2 = 64.
To write the equation of a circle centered at the point (–9, 3) with a radius of 8, we can use the general equation for a circle:
(x - h)^2 + (y - k)^2 = r^2
Where (h, k) represents the coordinates of the center of the circle, and r is the radius.
Substituting the given values into the equation, we get:
(x - (-9))^2 + (y - 3)^2 = 8^2
Simplifying further, we have:
(x + 9)^2 + (y - 3)^2 = 64
Therefore, the equation of the circle is (x + 9)^2 + (y - 3)^2 = 64.
recall that the circle with center at (h,k) and radius r is
(x-h)^2 + (y-k)^2 = r^2
so plug in your numbers.