QuestionDetermine the center and radius of the circle given by $x^{2}+12x+y^{2}-4y+15=0$ . Choose the correct option.

Studdy Solution

STEP 1

Assumptions1. The equation given is of a circle in the $x y$ -plane. . The general form of a circle's equation is $(x-h)^{}+(y-k)^{}=r^{}$ , where $(h,k)$ is the center of the circle and $r$ is the radius.

STEP 2

We need to rewrite the given equation in the form $(x-h)^{2}+(y-k)^{2}=r^{2}$ . To do this, we first group the $x$ terms and $y$ terms together.
$x^{2}+12 x+y^{2}-4 y+15=0$ can be rewritten as $(x^{2}+12 x) + (y^{2}-4 y) +15 =0$

STEP 3

Next, we complete the square for the $x$ terms and the $y$ terms.For the $x$ terms, we take half of the coefficient of $x$ , square it, and add it to the equation. The coefficient of $x$ is $12$ , so half of it is $6$ , and squaring it gives $36$ .For the $y$ terms, we take half of the coefficient of $y$ , square it, and add it to the equation. The coefficient of $y$ is $-$ , so half of it is $-2$ , and squaring it gives $$.We add these to the equation, remembering to subtract them from the other side to keep the equation balanced.

STEP 4

Rewrite the equation as $(x^{2}+12 x +36) + (y^{2}-4 y +4) =36 +4 -15$

STEP 5

The left side of the equation can now be rewritten as two squares, and the right side can be simplified $(x+)^{2} + (y-2)^{2} =25$

STEP 6

Now, we can see that the equation is in the form $(x-h)^{2}+(y-k)^{2}=r^{2}$ , so we can identify the center $(h,k)$ and the radius $r$ of the circle.The center is $(-6,2)$ and the radius is $\sqrt{25}=5$ .
So, the correct answer is A) center $(-6,2)$ , radius $=5$ .

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