Chapter 11.1, Problem 70STP

To determine

Write each equation $x^2+y^2+6y+13=40$ in standard form. Then graph the equation.

Answer to Problem 70STP

Equation in Standard form: $x^2+{(y+3)}^2=36$

Explanation of Solution

Given:

Write each equation in standard form. State whether the graph of the equation is a parabola, circle, ellipse or hyperbola. Then graph the equation.

The equation: $x^2+y^2+6y+13=40$

Concept Used:

The equation of circle in standard form: $x^2+y^2=r^2$

r is the radius of the circle.

Calculation:

The equation of circle in standard form: $x^2+y^2=r^2$

r is the radius of the circle.

$\begin{array}{}$ x^2+y^2+6y+13=40$$ x^2+\left(y^2+6y+9\right)+4=40$$ x^2+{(y+3)}^2+4-4=40-4$$ x^2+{(y+3)}^2=36$\end{array}$   $\begin{array}{l}\text{Center}:(0,-3)\\ \text{Radius}=6\end{array}$

Thus, the standard form of equation is$x^2+{(y+3)}^2=36$