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=0must show work thank you for the help

Expert Answers
jeew-m eNotes educator| Certified Educator


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+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.