Question

Write the equation in standard form. Identity the center and radius.
x² + y2 + 8x-4y-7=0

Answer 1

equation;

`(x + 4) {}^(2) + (y - 2) {}^(2) = 27`

Center (-4,2)

Radius is

`3 √(3)`

Explanation:

Since the x^2 and y^2 have the same coeffiecent this will be a circle in a form of

`(x - h) {}^(2) + (y - k) {}^(2) = {r}^(2)`

Where (h,k) is center

r is the radius

So first we group like Terms together

`{x}^(2) + 8x + {y}^(2) - 4y - 7 = 0`

Add 7 to both sides

`{x}^(2) + 8x + {y}^(2) - 4y = 7`

`( {x}^(2) + 8x) +( {y}^(2) - 4y) = 7`

Since the orginal form of the equation of the circle has a perfect square we need to complete the square for each problem

`((8)/(2) ) {}^(2) = 16`

and

`( - (4)/(2) ) {}^(2) = 4`

so we have

`{x}^(2) + 8x + 16 + {y}^(2) - 4y + 4y = 7 + 16 + 4`

`{x}^(2) + 8x + 16 + {y}^(2) - 4y + 4y = 27`

`(x + 4) {}^(2) + (y - 2) {}^(2) = 27`

To find our center, h is -4 and k is 2

so the center is (-4,2)

The radius is

`√(27) = 3 √(3)`

So the radius is 3 times sqr root of 3.