185k views
1 vote
Write the equation in standard form. Identity the center and radius.
x² + y2 + 8x-4y-7=0

Write the equation in standard form. Identity the center and radius. x² + y2 + 8x-example-1
User Cardin
by
7.8k points

1 Answer

10 votes

Answer:

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.

User Michael Romrell
by
8.0k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.