Respuesta :
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:
[tex]\large\boxed{x^2+y^2+4x-2y+1=0}[/tex]
Step-by-step explanation:
The standard form of an equation of a circle:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
(h, k) - center
r - radius
The general form of an equation of a circle:
[tex]x^2+y^2+Dx+Ey+F=0[/tex]
We have the center (-2, 1). Substitute to the equation in the standard form:
[tex](x-(-2))^2+(y-1)^2=r^2\\\\(x+2)^2+(y-1)^2=r^2[/tex]
Put thr coordinates of the point (-4, 1) to the equation and calculate a radius:
[tex](-4+2)^2+(1-1)^2=r^2\\\\r^2=(-2)^2+0^2\\\\r^2=4[/tex]
Therefore we have the equation:
[tex](x+2)^2+(y-1)^2=4[/tex]
Convert to the general form.
Use [tex](a\pm b)^2=a^2\pm 2ab+b^2[/tex]
[tex](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[/tex]
