MATH SOLVE

3 months ago

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

Accepted Solution

A:

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) - centerr - radiusThe 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]