x^2+y^2+2x+4y-31=0

Question

directions: write the standard form of the equation by completing the square&identify the conic section

Answer 1

Given equation is x^2+y^2+2x+4y-31 = 0

x^2+y^2+2x+4y-36+5 = 0

x^2+y^2+2x+4y+4+1-36 = 0

x^2+2x+1+y^2+4y+4-36 = 0

(x+1)^2+(y+2)^2-36 = 0

Add 36 to each side.

(x+1)^2+(y+2)^2-36+36 = 0+36

(x+1)^2+(y+2)^2 = 36

(x-(-1))^2+(y-(-4))^2 = 6^2

Compare it to standard form of crcle (x-h)^2+(y-k)^2 = r^2

(h,k) = (-2,-4)

radius is 6.

Comments

The above solution is wrong....

See the correct solution, answered by the lilly.....

Answer 2

The equation is x² + y² + 2x + 4y - 31 = 0.

To change the expression into a perfect square  add (half the x coefficient)² and add (half the y coefficient)²to each side of the expression.

Here, x coefficient = 2, so, (half the x coefficient)² = (2/2)²= 1.

Here y coefficient = 4, so, (half the y coefficient)² = (4/2)²= 4.

Add 1 and 4 to each side.

x² + 2x + 1 + y² + 4y + 4 - 31 = 0 + 1 + 4

(x + 1)² + (y + 2)² = 5 + 31 = 36

(x -  (- 1))² + (y - (- 2))² = 6².

Compare the above equation with the standard form of the circle equation : ( x - h )² + ( y - k )² = r², where, (h, k) is the center of the circle, and r is the radius.

The center (h, k) is (- 1, - 2), and the radius (r) is 6 units.

So, the equation  (x + 1)² + (y + 2)² = 6 represents a circle.