A circle with center (xo, yo) and radius r has an equation of the form:
(x - xo)2 + (y - yo)2 = r2
which can be rearranged to:
x2 - 2xox + xo2 + y2 - 2yoy + yo2 = r2
⇒ x2 - 2xox +y2 - 2yoy + xo2 + yo2 - r2 = 0
or x2 - 2xox +y2 - 2yoy = r2 - xo2 - yo2
So depending how your equation is given, the center can be read directly from the equation (first form above) or is -1/2 times the coefficients of the x and y terms. eg:
What is the center of the circle with equation x2 - 4x + y2 + 6y + 12 = 0
The coefficients of the x and y terms are -4 and +6, so the center is -1/2 times these, namely; (2, -3).
Fully rearranging the circle's equation:
x2 - 4x + y2 + 6y + 12 = 0
⇒ (x - 2)2 - 4 + (y + 3)2 - 9 + 12 = 0 [completing the squares]
⇒ (x - 2)2 + (y + 3)2 - 1 = 0
⇒ (x - 2)2 + (y + 3)2 = 1
⇒ circle has centre (2, -3) and radius 1.
Equation of the circle: (x-3)^2 +( y+13)^2 = 169
There are probably several ways to approach it; one general equation for the circle is: (x - a)2 + (y - b)2 = r2 This describes a circle with center at coordinates (a, b), and with a radius of r.
to be honest I dont know
x2+y2+4x+2y+3=0(x+2)2 + (Y+2.5)2 = 3This is the equation of circle with center at (-2,-2.5) with radius 3.5
The equation of a circle with center (0,2) and radius r is x^2+(y-2)^2=r^2 Since it passes through (0,0) (the origin) 0^2+(0-2)^2=r^2 r^2=4 The equation of the circle is x^2+(y-2)^2=4
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.
a diameter
The center of the circle is at (9, 7) on the Cartesian plane
Equation of the circle: (x-3)^2 +( y+13)^2 = 169
To find the standard equation for a circle centered at the origin, we use the distance formula to define the radius. The equation is derived from the relationship that the distance from any point ((x, y)) on the circle to the center ((0, 0)) is equal to the radius (r). Thus, the standard equation of the circle is given by (x^2 + y^2 = r^2). Here, (r) is the radius of the circle.
(3, -2) You find the center by taking the opposite of the numbers added to the X and to the Y.
multiple the diameter by 2 to get the radius, then use your equation and go from there
There are probably several ways to approach it; one general equation for the circle is: (x - a)2 + (y - b)2 = r2 This describes a circle with center at coordinates (a, b), and with a radius of r.
to be honest I dont know
To find the equation of a circle with center ((2, -1)) that passes through the point ((3, 4)), we first calculate the radius. The radius (r) is the distance between the center and the point, which is given by (r = \sqrt{(3 - 2)^2 + (4 - (-1))^2} = \sqrt{1^2 + 5^2} = \sqrt{26}). The standard equation of a circle is ((x - h)^2 + (y - k)^2 = r^2), where ((h, k)) is the center. Thus, the equation is ((x - 2)^2 + (y + 1)^2 = 26).
To draw a flowchart for finding the equation of a circle passing through three given points, start by defining the three points as ( A(x_1, y_1) ), ( B(x_2, y_2) ), and ( C(x_3, y_3) ). Next, set up the general equation of a circle ( (x - h)^2 + (y - k)^2 = r^2 ) and derive a system of equations by substituting the coordinates of the points into this equation. Solve the resulting system of equations for the center coordinates ( (h, k) ) and the radius ( r ), and finally, express the equation of the circle in standard form.
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})).