find the x and y intercepts for x^2+y^2+4x+2y-20=0

Question

I need to find the x and y intercepts of :

x^2+y^2+4x+2y-20=0.

Answer 1

x^2+y^2+4x+2y-20 = 0

It is a circle equation.

To find x intercept substitute y = 0 in given equation.

x^2+4x-20 = 0

Compare it to quadratic form ax^2+bx+c = 0

a = 1, b = 4, c = -20.

x = [-b±√(b^2-4ac)]/2a

x = [-4±√(16+80)]/2

x = [-4±√96]/2

x = [-4±9.79]/2

x = (-4+9.79)/2 and x = (-4-9.79)/2

x = 5.79/2 and x = -13.79/2

x = 2.895 and x = -6.895

x = 2.9 and x = -6.9

To find y intercept substitute x = 0 in given equation.

y^2+2y-20 = 0

Compare it to quadratic form ax^2+bx+c = 0

a = 1, b = 2, c = -20.

y = [-b±√(b^2-4ac)]/2a

y = [-2±√(4+80)]/2

y = [-2±9.1]/2

y = (-2+9.1)/2 and y = (-2-9.1)/2

y = 7.1/2 and y = -11.1/2

y = 3.55 and y = -5.55

X intercepts are 2.9 and -6.9 & y intercepts are 3.55 and -5.55.