# Graph the following equation and state the center, radius and x- and y- intercepts if they exist. `(x+1)^2+(y-6)^2=36` Please show all your work.Need correct answer by today

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### 1 Answer

This question asks you to find some information about a circle given the equation:

`(x+1)^2 + (y-6)^2 = 36`

Thankfully, this equation is almost in the form we need to find all of the information the question asks for.

**Finding the center and radius**

Here is the standard equation of a circle (see link below):

`(x-a)^2+(y-b)^2 = r^2`

Where `(a,b)` is the center of the circle and `r^2` is the square of the radius.

We just need to change one sign around and get 36 into the form 6^2 in order to get the form in the above formula, and we should have our center and radius:

`(x-(-1))^2 + (y-6)^2 = 6^2`

There it is, our center will be (-1, 6) and our radius is 6.

**Finding the x and y intercepts**

To find these intercepts, you must recall how the intercepts are defined. At an x-intercept, y = 0. At a y-intercept, x = 0.

To find the **x-intercepts**, we set y = 0 in our equation and solve for the x-values that allow y to be 0:

`(x+1)^2 + (0 - 6)^2 = 36`

Now, let's simplify by squaring the terms in the parentheses and combining "like" terms:

`x^2+2x+1+36 = 36`

`x^2+2x+37 = 36`` `

Now, we subtract 36 from both sides to get a quadratic equation:

`x^2+2x+1 = 0`

We could use the quadratic equation to find the x-values, but it is much easier to factor as we do below:

`(x+1)(x+1) = 0`

To solve this equation, we simply allow one or the other expression inside the parentheses to be equal to 0. This particular case is easy because both expressions are the same!

`x+1 =0`

Solving by subtracting 1 from both sides:

`x = -1`

This gives us our only **x-intercept** (keeping in mind from before that y = 0):

**(-1, 0)**

To find the y-intercepts, we'll do pretty much the same thing. The difference is that where we intersect the y-axis, we have an x-value of 0, so we'll need to solve for the y-values. Let's put x = 0 into our equation:

`(0+1)^2 + (y-6)^2 = 36`

Now, we'll simplify again:

`1 + y^2-12y+36 = 36`

Again, we'll subtract 36 from both sides:

`y^2-12y+1 = 0`

Now, we have a different quadratic equation, which we'll have some trouble solving through factoring, so we'll just use the quadratic equation:

`y = (-(-12) +- sqrt((-12)^2 - 4(1)(1)))/(2(1))`

Evaluating this gives us the following 2 values for y:

y = 0.084 and y = 11.92

This gives us our **y-intercepts**:

**(0, 0.084) and (0, 11.92)**

**As a summary:**

Equation:

`(x+1)^2+(y-6)^2=36`

Radius:

`r=6`

Center

`C = (-1, 6)`

X-Intercept:

(-1, 0)

Y-Intercepts:

(0, 0.084), (0, 11.92)

To see if these values are reasonable, let's graph the equation:

Believe it or not, that's a circle. The scale is just a bit off. However, if you make estimates based on the values we determined above, you'll see that they all fit pretty well with this graph. Hope that helps!

**Sources:**