Question

what is the general form of the equation of a circle with its Center at -2 1 and passing through -4 one

Answer 1

Answer:

x² + y² + 4x - 2y + 1 = 0

Step-by-step explanation:

The equation of a circle is given by the general equation;

(x-a)² + (y-b)² = r² ; where (a,b) is the center of the circle and r is the radius.

In this case; the center is (-2,1)

We can get radius using the formula for magnitude; √((x2-x1)² + (y2-y1)²)

Radius = √((-4- (-2))² + (1-1)²)

             = 2

Therefore;

The equation of the circle will be;

(x+2)² + (y-1)² = 2²

(x+2)² + (y-1)² = 4

Expanding the equation;

x² + 4x + 4 + y² -2y + 1 = 4 subtracting 4 from both sides;

x² + 4x + y² - 2y + 4 + 1 -4 = 0

= x² + y² + 4x - 2y + 1 = 0

Answer 2

Answer:

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

Step-by-step explanation:

The standard form of an equation of a circle:

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

(h, k) - center

r - radius

The general form of an equation of a circle:

$x^2+y^2+Dx+Ey+F=0$

We have the center (-2, 1). Substitute to the equation in the standard form:

$(x-(-2))^2+(y-1)^2=r^2\\\\(x+2)^2+(y-1)^2=r^2$

Put thr coordinates of the point (-4, 1) to the equation and calculate a radius:

$(-4+2)^2+(1-1)^2=r^2\\\\r^2=(-2)^2+0^2\\\\r^2=4$

Therefore we have the equation:

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

Convert to the general form.

Use $(a\pm b)^2=a^2\pm 2ab+b^2$

$(x+2)^2+(y-1)^2=4\\\\x^2+2(x)(2)+2^2+y^2-2(y)(1)+1^2=4\\\\x^2+4x+4+y^2-2y+1=4\qquad\text{subtract 4 from both sides}\\\\x^2+y^2+4x-2y+1=0$