Answer:
equation;
[tex](x + 4) {}^{2} + (y - 2) {}^{2} = 27[/tex]
Center (-4,2)
Radius is
[tex]3 \sqrt{3} [/tex]
Step-by-step explanation:
Since the x^2 and y^2 have the same coeffiecent this will be a circle in a form of
[tex](x - h) {}^{2} + (y - k) {}^{2} = {r}^{2} [/tex]
Where (h,k) is center
r is the radius
So first we group like Terms together
[tex] {x}^{2} + 8x + {y}^{2} - 4y - 7 = 0[/tex]
Add 7 to both sides
[tex] {x}^{2} + 8x + {y}^{2} - 4y = 7[/tex]
[tex]( {x}^{2} + 8x) +( {y}^{2} - 4y) = 7[/tex]
Since the orginal form of the equation of the circle has a perfect square we need to complete the square for each problem
[tex] (\frac{8}{2} ) {}^{2} = 16[/tex]
and
[tex]( - \frac{4}{2} ) {}^{2} = 4[/tex]
so we have
[tex] {x}^{2} + 8x + 16 + {y}^{2} - 4y + 4y = 7 + 16 + 4[/tex]
[tex] {x}^{2} + 8x + 16 + {y}^{2} - 4y + 4y = 27[/tex]
[tex](x + 4) {}^{2} + (y - 2) {}^{2} = 27[/tex]
To find our center, h is -4 and k is 2
so the center is (-4,2)
The radius is
[tex] \sqrt{27} = 3 \sqrt{3} [/tex]
So the radius is 3 times sqr root of 3.
