Question

What is the general form of the equation of a circle with its center at (-2, 1) and passing through (-4, 1)?

Answer 1

The general form of the equation of a circle with center (-2,-1) and passing through (-4,1) is given by;

$x^{2} +y^{2}+4x-2y+1=0$

Further Explanation:

Equation of circle

• An equation of a circle is calculated when the radius of a circle and the coordinates of its center are known or can be calculated.
• Equation of a circle can be written in two forms namely; the standard and the general form.

Standard form

• Standard form of Equation of a circle is written as;

$(x-a)^{2} +(y-b)^{2}= r^{2}$

Where (a,b) is the center of the circle and r is the radius of the circle.

• When an equation is written in this form the center and the radius can easily be identified from the equation, with radius always being positive.

General form

• The general form of an equation of a circle is given by;

$x^{2} +y^{2} + dx+ey+f=0$ ; where d, e and f are constants.

• When an equation is written in this form one has to use the completing square method to write it in standard form so as to identify the center and the radius.

In this case;

Center of the circle (a,b) is (-2,1)

To get the radius of the circle we use magnitude;

square root of( (x1-x2)^2 + (y1-y2)^2) where X1 = -2, x2= -4, while y1= 1 and y2 =1

The radius will be; 4 units

Therefore;  Using the equation

$(x-a)^{2} +(y-b)^{2}= r^{2}$

Substituting the values in the general equation we get;

$(x+4)^{2} +(y-1)^{2}= 4^{2}$

Expanding the equation we get;

$x^{2} +y^{2}+4x-2y+1=0$ , which is the general form of the equation of the circle.

Keywords: Circle, equation of the circle, center, radius

Learn more about:

• Equation of a circle; brainly.com/question/1493255
• Standard form of equation of a circle:brainly.com/question/1493255
• General form of equation of a circle; brainly.com/question/1493255

Level: High school

Subject: Mathematics

Topic: Equation of the circle

Answer 2

Equation is :
( x + 2 )² + ( y - 1 )² = r²
( - 4 + 2 )² + ( 1 - 1 )² = r²
4 = r², than we will plug in to a formula:
( x + 2 )² + ( y - 1 )² = 4
x² + 4 x + 4 + y² - 2 y +1 = 4
x² + y² + 4 x - 2 y + 1 = 0
Answer: B )