Question

A circle has the equation 2x²+12x+2y²−16y−150=0.

Answer 1

Answer: B. The coordinates of the center are (-3,4), and the length of the radius is 10 units.

Step-by-step explanation:

The equation of a circle in the center-radius form is:

$(x-h)^{2} +(y-k)^{2}=r^{2}$ (1)

Where $(h,k)$ are the coordinates of the center and $r$ is the radius.

Now, we are given the equation of this circle as follows:

$2x^{2}+12x+2y^{2}-16y-150=0$ (2)

And we have to write it in the format of equation (1). So, let's begin by applying common factor 2 in the left side of the equation:

$2(x^{2}+6x+y^{2}-8y-75)=0$ (3)

Rearranging the equation:

$x^{2}+6x+y^{2}-8y=75$ (4)

$(x^{2}+6x)+(y^{2}-8y)=75$ (5)

Now we have to complete the square in both parenthesis, in order to have a perfect square trinomial in the form of $(a\pm b)^{2}=a^{2}\pm+2ab+b^{2}$:

For the first parenthesis:

$x^{2}+6x+b^{2}$

We can rewrite this as:

$x^{2}+2(3)x+b^{2}$

Hence in this case $b=3$ and $b^{2}=9$:

$x^{2}+2(3)x+3^{2}=x^{2}+6x+9=(x+3)^{2}$

For the second parenthesis:

$y^{2}-8y+b^{2}$

We can rewrite this as:

$y^{2}-2(4)y+b^{2}$

Hence in this case $b=-3$ and $b^{2}=9$:

$y^{2}-2(4)y+4^{2}=y^{2}-8y+16=(y-4)^{2}$

Then, equation (5) is rewritten as follows:

$(x^{2}+6x+9)+(y^{2}-8y+16)=75+9+16$ (6)

Note we are adding 9 and 16 in both sides of the equation in order to keep the equality.

Rearranging:

$(x-3)^{2}+(y-4)^{2}=100$ (7)

At this point we have the circle equation in the center radius form $(x-h)^{2} +(y-k)^{2}=r^{2}$

Hence:

$h=-3$

$k=4$

$r=\sqrt{100}=10$