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lara [203]
2 years ago
9

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

Mathematics
1 answer:
lorasvet [3.4K]2 years ago
8 0

Answer:

equation;

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

Center (-4,2)

Radius is

3 \sqrt{3}

Step-by-step 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

(\frac{8}{2} ) {}^{2}  = 16

and

( - \frac{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

\sqrt{27}  = 3 \sqrt{3}

So the radius is 3 times sqr root of 3.

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