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Stella [2.4K]
3 years ago
12

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

Mathematics
1 answer:
KonstantinChe [14]3 years ago
8 0

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}:

<u>For the first parenthesis:</u>

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}

<u>For the second parenthesis:</u>

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)

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

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

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