0
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.
What else can I help you with?
The center of a circle is located at c(3-13). The x-axis is tangent to the circle at (30). Find the equation of the circle?
Equation of the circle: (x-3)^2 +( y+13)^2 = 169
How do you find the general form equation of a circle?
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.
How do you find the center or midpoint of a circle given one endpoint of the diameter?
to be honest I dont know
Find the center and radius of the circle with this equation x2 plus y2 plus x equals 0?
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
Find an equation of the circle that has center 0 2 and passes through the origin.?
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
What is the center of the circle given by the equation (x - 3)2 (y - 9)2 16?
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.
What is a chord passing through the center of a circle?
a diameter
Find the center of a circle with the equation x2 plus y2-18x-14y plus 124?
The center of the circle is at (9, 7) on the Cartesian plane
The center of a circle is located at c(3-13). The x-axis is tangent to the circle at (30). Find the equation of the circle?
Equation of the circle: (x-3)^2 +( y+13)^2 = 169
How To find the standard equation for a circle centered at the origin we use the distance formula since the radius measures?
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.
Where is the center of the circle given by the equation x - 3 2 plus y plus 2 2 equals 9?
(3, -2) You find the center by taking the opposite of the numbers added to the X and to the Y.
How do you find the area of a circle if the diameter is given?
multiple the diameter by 2 to get the radius, then use your equation and go from there
How do you find the general form equation of a circle?
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.
How do you find the center or midpoint of a circle given one endpoint of the diameter?
to be honest I dont know
What is the center and radius of a circle described by the equation x2 y2 8x 4y 16 0?
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})).
How to find the center of a circle in woodworking?
To find the center of a circle in woodworking, draw two diagonal lines from opposite corners of the circle. Where the lines intersect is the center of the circle.
Find the center and radius of the circle with this equation x2 plus y2 plus x equals 0?
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
