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Standard Form of the Equation of a Circle with Center at the Origin and Radius 9

In a rectangular coordinate plane where the center of a circle is of radius r (h,k), we have The center is or, in other words. If you replace and for , our formula becomes the general form of the equation of the circle with center ((x_1, y_1)) and radius (r) is ( x^2 + y^2 + Ax + By + C = 0), where (A = -2x_1) (B = -2y_1) (C = {x_1}^2 + {y_1}^2 -r^2) Since the origin is (0,0), we know, that h and k are both 0. Since they give us the radius, we also know that r=9 Radius r = (sqrt{g^2+f^2 – c}) = (sqrt{(-3)^{2}+(-4)^{2} – 9}) = (sqrt{9 + 16 – 9}) = (sqrt{16}) = 4th radius r = 4. This is the standard equation of the circle, with a radius r and a center at (a,b): (x – a)2 + (y – b)2 = r2 and think of the general form as: x2 + y2 + 2gx + 2fy + c = 0. Here are the steps to convert the general shape into a standard shape: the circle has a center at (5.5) and a radius of 2. Therefore, equation (x-5)2+(y-5)2=22 or (x-5)2+(y-5)2=4. General equation of a circle: #(x-a)^2+(y-b)^2=r^2#, where (a,b) are the coordinates of the center and `r` is the radius, since in your question the center is on the origin and the radius is 9, so (a,b) = (0,0) and r=9 so #x^2 + y^2 = 9^2# Which of the following equations describes all the points (x, y) in a coordinate plane five units from the point (–3, 6)? Here is a list of some points to keep in mind when studying the equation of the circle Now that we have the new center and radius of the circle A, we can write its general equation with (x – h)2 + (y – k)2 = r2. The general form of the circular equation is: x2 + y2 + 2gx + 2fy + c = 0. This general shape is used to find the coordinates of the center of the circle and the radius of the circle.

Here, c is a constant term, and the equation with a value c represents a circle that does not pass through the origin. Consider the case where the center of the circle is on the x-axis: (a, 0) is the center of the circle with the radius r. (x, y) is any point on the circumference of the circle. Replace h, k, and r to find the equation in standard form. Since (h,k)=(−1,0) and r=3 we have, the general form of the equation of a circle is: x2 + y2 + 2gx + 2fy + c = 0. This general shape is used to find the coordinates of the center of the circle and the radius, where g, f, c are constants. Unlike the standard form, which is easier to understand, the general shape of a circle`s equation makes it difficult to find significant properties on a particular circle. So we will complete the square formula to make a quick conversion from the general shape to the standard form. (rcosθ)2 + (rsinθ)2 = p2 r2cos2θ + r2sin2θ = p2 r2(cos2θ + sin2θ) = p2 r2(1) = p2 r = p where p is the radius of the circle. (x – h)2 + (y – k)2 = r2, where the center is at (h, k) and r is the length of the radius.

If we know the coordinates of the center of a circle and the radius, we can find the general equation of the circle. For example, the center of the circle (1, 1) and the radius is 2 units, then the general equation of the circle can be obtained by replacing the center and radius values. The general equation of the circle is (x^2 + y^2 + Ax + By + C = 0). To determine the radius of a circle, you must take the number that corresponds to the equation and assign it the square root root. This is due to the square above. Use the less common multiples of 27 to discover that three make up 3 27. Take out two trios because the square root of a number multiplied by itself is itself. This leaves a 3 under the radical.

Therefore, our radius. Let`s take both ends of the diameter like (1, 1) and (3, 3). First, calculate the center point using the section formula. The coordinates of the center will be (2, 2). Second, calculate the radius formula by distance between (1, 1) and (2, 2). The radius is (sqrt{2}). Now the equation of the circle in standard form is ({(x – 2)}^2 + {(y – 2)}^2 = 2). (x – 2)2 + (y – (-3))2 = (3)2 (x – 2)2 + (y + 3)2 = 9 is the required standard form of the equation of the given circle. In this form, the center and radius are obvious. For example, given the equation (x−2) 2 + (y + 5) 2 = 16 that we have, consider the case where the circumference of the circle touches the x-axis at a certain point: (a, r) is the center of the circle of radius r. If a circle touches the x-axis, then the y-coordinate of the center of the circle is equal to the radius r.

The standard equation of a circle gives accurate information about the center of the circle and its radius and so it is much easier to read the center and radius of the circle at a glance. .

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