write the equation of the circle in standard form. find the center, radius, intercepts and graph the circle. x^2+y^2+10x+8y+16=0 must show work thank you for the help
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The standard form of notation for a circle is;
`(x-h)^2+(y-k)^2 = r^2`
Here (h,k) is the center point cordinates and r is the radius of the circle.
Lets consider (x+a)^2 and (x-a)^2
`(x+a)^2 = x^2+2ax+a^2`
`(x-a)^2 = x^2-2ax+a^2`
When we consider the x component at right side it is twice the product of x*a with same sign between x and a in left side.
This fact is used now onwards in the solution.
`x^2+y^2+10x+8y+16=0`
`(x+5)^2-25 = x^2+10x`
`(y+4)^2-16 = y^2+8x`
`(x+5)^2-25+(y+4)^2-16+16 = 0`
`(x+5)^2+(y+4)^2 = 25`
`(x+5)^2+(y+4)^2 = 5^2`
So the equation is in the standard form.
The cordinates of the center is (-5,-4) and the radius of the circle is 5.
When x = 0 then;
`y^2+8y+16 =0`
`y^2+8y+16 = 0 `
`(y+4) = 0 `
`y =-4`
So the grapgh intercept x-axis at y=-4
When y = 0 then;
`x^2+10x+16 = 0`
`x^2+8x+2x+16 = 0`
`(x+8)(x+2) = 0`
`x = -8 or x = -2`
So the circle intercept y-axis at x=-8 and x=-2
The graph is shown below.
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