Question

The equation of a circle is x^2+y^2+18x+4y+49=0. What are the center and the radius of the circle?

please help I am failing math and have no idea what's :) going :) on :) in :) class :)

Answer 1

Center: (-9, -2)

Radius = 6

Explanation:

The general equation of the circle is:

`x^(2) + y^(2)+2gx+2fy+c=0`

The center of the circle is given as (-g, -f) and the radius of this circle is calculated as:

`r=\sqrt{g^(2)+f^(2)-c}`

The given equation is:

`x^(2) +y^(2)+18x+4y+49=0`

Re-writing this equation in a form similar to general form:

`x^(2) +y^(2)+2(9)(x)+2(2)(y)+49=0`

Comparing this equation with general equation we get:

g = 9

f = 2

c = 49

Thus center of the given circle is (-g, -f) = (-9, -2)

The radius of the circle will be:

`r=\sqrt{9^(2)+2^(2)-49}=6`

Thus the radius of the given circle is 6.