Question

for the equation x^2+y^2-6x+4y=121. find the center2. graph the circle3. find the intercepts (radical form)

Answer 1

ANSWER and EXPLANATION

1. We want to find the center of the given equation:

`x^2+y^2-6x+4y=12`

To do this, complete the square for the two variables in the equation, x and y:

`\begin{gathered} x^2-6x+y^2+4y=12 \\ \\ x^2-6x+(-(6)/(2))^2+y^2+4y+((4)/(2))^2=12+(-(6)/(2))^2+((4)/(2))^2 \\ \\ x^2-6x+9+y^2+4y+4=25 \end{gathered}`

Now, factorize the x and y parts of the equation and write the equation of the circle in general form:

`\begin{gathered} (x^2-3x-3x+9)+(y^2+2y+2y+4)=25 \\ \\ x(x-3)-3(x-3)+y(y+2)+2(y+2)=5^2 \\ \\ (x-3)^2+(y+2)^2=5^2 \end{gathered}`

In the general form of the equation of a circle:

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

the center of the circle is (h, k).

Therefore, comparing the given equation to the general equation of a circle, the center of the circle is:

`(3,-2)`

2. Now, we want to graph the circle. The graph of the circle is given below:

3. To find the x and y-intercepts of the circle, we have to find the points where the graph of the circle touches the x and y axes.

For the y-intercepts:

`(0,-6)\text{ and }(0,2)`

That is the answer.

Since we want to write the x-intercepts in radical form, we can solve for them from the given equation.

To find the x-intercepts, solve for x when y is equal to 0:

`\begin{gathered} x^2+0^2-6x+(0)y=12 \\ \\ x^2-6x-12=0 \end{gathered}`

Solve using the quadratic formula:

`x=(-b\pm√(b^2-4ac))/(2a)`

where a = 1, b = -6, c = -12

Therefore, the x-intercepts are:

`\begin{gathered} x=(-(-6)\pm√((-6)^2-4(1)(-12)))/(2(1))=(6\pm√(36+48))/(2) \\ \\ x=(6\pm√(84))/(2)=(6\pm2√(21))/(2) \\ \\ x=3+√(21)\text{ and }x=3-√(21) \end{gathered}`

Those are the x-intercepts in radical form.