Question

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

A.
x2 + y2 − 4x + 2y + 1 = 0

B.
x2 + y2 + 4x − 2y + 1 = 0

C.
x2 + y2 + 4x − 2y + 9 = 0

D.
x2 − y2 + 2x + y + 1 = 0

Answer 1

B. x² + y² + 4x − 2y + 1 = 0

Explanation:

The standard form of an equation of a circle:

`(x-h)^2+(y-k)^2=r^2`

(h, k) - center

r - radius

The formula of a distance between two points:

`d=√((x_2-x_1)^2+(y_2-y_1)^2)`

We have the center (-2, 1) → h = -2 and k = 1.

The radius is the distance between the center and the given point.

Therefore put the coordinates of the points (-2, 1) and (-4, 1) to the formula of a distance between two points:

`r=√((-4-(-2))^2+(1-1)^2)=√((-2)^2+0^2)=√(4)=2`

We have the equation in standard form:

`(x-(-2))^2+(y-1)^2=2^2\\\\(x+2)^2+(y-1)^2=4`

Convert to the genereal form using:

`(a\pm b)^2=a^2\pm2ab+b^2`

`(x+2)^2+(y-1)^2=4\\\\x^2+2(x)(2)+2^2+y^2-2(y)(1)+1^2=4\\\\x^2+4x+4+y^2-2y+1=4\\\\x^2+y^2+4x-2y+5=4\qquad\text{subtract 4 from both sides}\\\\\boxed{x^2+y^2+4x-2y+1=0}\to\boxed{B.}`