Example Question - circle equation
Here are examples of questions we've helped users solve.
Finding the Diameter of a Circle from a Given Equation
Complete the square for \( x \) and \( y \): \( x^2 + 22x = x^2 + 22x + (22/2)^2 - (22/2)^2 = (x+11)^2 - 121 \) \( y^2 + 30y = y^2 + 30y + (30/2)^2 - (30/2)^2 = (y+15)^2 - 225 \) Rewrite the equation: \( (x+11)^2 - 121 + (y+15)^2 - 225 + 90 = 0 \) \( (x+11)^2 + (y+15)^2 = 121 + 225 - 90 \) \( (x+11)^2 + (y+15)^2 = 256 \) The equation \( (x+11)^2 + (y+15)^2 = 256 \) represents a circle with radius \( r^2 = 256 \), so \( r = 16 \). Diameter \( d = 2r = 2 \times 16 = 32 \). The diameter of the circle is 32.
Solving for Radius of a Circle Using Completing the Square Method
Para resolver la pregunta en la imagen, la cual pide encontrar el radio de una circunferencia, debemos completar el cuadrado para los términos con \( x \) e \( y \) de la ecuación dada: \( x^2 + y^2 + 4x + 4y - 17 = 0 \) Primero, vamos a reorganizar los términos y agruparlos por variable: \( x^2 + 4x + y^2 + 4y = 17 \) Ahora, para completar el cuadrado, debemos encontrar dos números que al elevarlos al cuadrado y sumarlos a los términos en \( x \) e \( y \) nos permitan formar dos cuadrados perfectos. Para \( x \), el número es \( (4/2)^2 = 4 \), y para \( y \), es \( (4/2)^2 = 4 \). Sumamos estos números a ambos lados de la ecuación: \( x^2 + 4x + 4 + y^2 + 4y + 4 = 17 + 4 + 4 \) Al sumarlos nos queda: \( (x + 2)^2 + (y + 2)^2 = 25 \) La ecuación de una circunferencia con centro en \( (h, k) \) y radio \( r \) es \( (x - h)^2 + (y - k)^2 = r^2 \). Comparando esta forma con nuestra ecuación: \( (x + 2)^2 + (y + 2)^2 = 25 \) Podemos ver que el centro de la circunferencia es \( (-2, -2) \) y que \( r^2 = 25 \). Tomando la raíz cuadrada de ambos lados de la ecuación \( r^2 = 25 \): \( r = \sqrt{25} \) \( r = 5 \) Por lo tanto, el radio de la circunferencia es 5.
Circle Equation with Center and Radius
The equation given in the image is for a circle, and it is written in the standard form: \[ (x - h)^2 + (y - k)^2 = r^2 \] where \((h, k)\) is the center of the circle, and \(r\) is the radius. The given equation is: \[ (x - 3)^2 + (y - 1)^2 = 4 \] From this equation, we can directly read the center and the radius of the circle: Center \((h, k)\) is \((3, 1)\), as it's the point you get by undoing the sign change in the brackets. Radius \(r\) is \(\sqrt{4}\), which is \(2\), as the radius squared \(r^2\) equals \(4\). So the center of the circle is at the point \((3, 1)\), and the radius is \(2\). To represent this graphically, you would plot the center at point \((3, 1)\) on a Cartesian plane and draw a circle around this point with a radius of \(2\) units, ensuring that all points on the circumference of the circle are \(2\) units away from the center.
Equation of a Circle with Center and Radius
The question is asking for the equation of the circle with a center at point (2, 0) and a radius of 2. It also requests that the circle be drawn. The equation of a circle in the coordinate plane with a center at (h, k) and radius r is given by: \[ (x - h)^2 + (y - k)^2 = r^2 \] For this circle: - h = 2 - k = 0 - r = 2 So plugging these values into the equation, we get: \[ (x - 2)^2 + (y - 0)^2 = 2^2 \] \[ (x - 2)^2 + y^2 = 4 \] This is the equation of the circle with a center at (2, 0) and a radius of 2. To draw this circle, you would plot the center at (2, 0) and use a compass set to a width of 2 units (the radius) to draw the circumference, ensuring it is a perfect circle around the center point.
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