x^2+y^2+6x-4y-23=0 convert to standard form

Question

convert to standard form.

Answer 1

x^2+y^2+6x-4y-23 = 0

x^2+6x+y^2-4y-23 = 0

Add 13 to each side.

x^2+6x+y^2-4y-23+13 = 13

x^2+6x+9+y^2-4y+4-23 = 13

(x+3)^2+(y-2)^2-23 = 13

Add 23 to each side.

(x-(-3))^2 + (y-2)^2 = 36

(x-(-3))^2 + (y-2)^2 = 6^2

Now it is in standard form of circle is (x-h)^2+(y-k)^2 = r^2.

Answer 2

Standard form of a circle equation is

Where center is (h ,k ) and radius of circle is r .

The equation is x ²  + y ²+ 6x - 4y - 23 = 0

x ² + 6x + y ² - 4y  = 23

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 = 6. so, (half the x coefficient)² = (6/2)²= 9

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

Add 4 and 9 to each side.

x ²  + 6x + 9 + y ² - 4y  + 4 = 23 + 4 + 9

(x + 3)²  + (y - 2)²  = 36

(x - (-3))²  + (y - 2)²  = 6²

Compare it to standard form of circle

Center of circle = (-3,2)

Radius of circle = 6.

Standard form of the circle is (x - (-3))²  + (y - 2)²  = 6² .