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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
Which is an equation of a circle with center (2 -1) that passes through the point (3 4)?
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).
How do you draw a flow chart to find the equation of a circle passing through three given points?
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.
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})).
