Respuesta :

gmany

The equation of a circle:

[tex](x-h)^2+(y-k)^2=r^2[/tex]

(h, k) - center

r - radius

Look at the picture. We have

center: (1, -2) → h = 1, k = -2

radius: r = 3

Substitute:

[tex](x-1)^2+(y-(-2))^2=3^2\\\\(x-1)^2+(y+2)^2=9\ \ \ |\text{use}\ (a\pm b)^2=a^2\pm2ab+b^2\\\\x^2-2(x)(1)+1^2+y^2+2(y)(2)+2^2=9\\\\x^2-2x+1+y^2+4y+4=9\ \ \ |\text{use commutative property}\\\\x^2+y^2-2x+4y+5=9\ \ \ \ |-9\\\\x^2+y^2-2x+4y-4=0\\\\Answer:\ a=-2,\ b=4,\ c=-4[/tex]

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abidemiokin

The value of b for the circle in general form is 4

The standard equation of a circle is expressed as:

[tex](x-a)^2+(y-b)^2=r^2[/tex]

From the given diagram, the centre of the circle is at (-1, 2) and the radius is 3 units

Substitute into the formula:

[tex](x-1)^2+(y+2)^2=3^2\\[/tex]

Expand the expression to have:

[tex]x^2-2x+1+y^2+4y+4 -9 =0\\x^2+y^2-2x+4y-4=0[/tex]

Compare the result with the given expression [tex]x^2+y^2+ax+by+c=0[/tex]

ax = -2x

a = -2

4y = by

b = 4

Hence the value of b for the circle in general form is 4

Learn more here: https://brainly.com/question/23226948

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