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...

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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.**