Question

The equation of a given circle in. general form is x^2+y^2+12y+27=0. Write the equation in standard form, (x-h)^2+(y-k)^2=r^2, by completing the squares in the equation. Show your work in the table.

Answer 1

Answer:

Center (h, k) = (0, 6)

Radius, r = 3 units

Step-by-step explanation:

The given equation of the circle is: $x^2 + y^2 + 12y + 27 = 0$

STEP I: Move the constant term to the right hand side.

To do that we subtract 27 on both the sides, we get:

$x^2 + y^2 + 12y = -27$

STEP II: Divide the co - effecient of 'x' by 2 and square it.

Here there is no 'x' term. Its co-effecient is 0.

STEP III: Add the result on both sides.

Since, we got 0 in the previous step, there would no change in the equation

STEP IV: We repeat the process for 'y'.

Here, the co-effecient of 'y' is: 12.

Dividing by 2 and squaring it will yield:

$\frac{12}{2} = 6; \hspace{5mm} 6^2 = 36$

STEP V: Adding 36 to both the sides of the equation, it becomes:

$x^2 + y^2 + 12y + 36 = -27 + 36$

$\implies x^2 + y^2 + 12y + 36 = 9$

We have arrived at perfect squares.

STEP VI: Write in the form of perfect squares.

$(x - 0)^2 + (y - 6)^2 = 3^2$

STEP VII: Compare it with the standard form.

$(x - h)^2 + (y - k)^2 = r^2$

We get: The centre of the circle: (h, k) = (0, 6) and

the radius of the circle, r = 3 units.