Chapter 13.6, Problem 16PSC

To determine

To find: The centre and the radius of the graph of the given equation.

Answer to Problem 16PSC

Thecentre is at $\left(-2,\frac{5}{6}\right)$ and radius is $\sqrt{\frac{193}{36}}$ .

Explanation of Solution

Given:

The given equation is $3x^2+12x+3y^2-5y=2$ .

Calculation:

Consider the formula for the equation of the circle is of the form as,

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Here, $\left(h,k\right)$ is the centre and $r$ is the radius.

Consider the given equation is solved as,

  $\begin{array}{l}3x^2+12x+3y^2-5y=2\\ x^2+4x+y^2-\frac{5}{3}y=\frac{2}{3}\\ x^2+4x+4+y^2-\frac{5}{3}y+\frac{25}{36}=\frac{2}{3}+4+\frac{25}{36}\\ {\left(x+2\right)}^2+{\left(y-\frac{5}{6}\right)}^2=\frac{193}{36}\end{array}$

Thus, the centre is at $\left(-2,\frac{5}{6}\right)$ and radius is $\sqrt{\frac{193}{36}}$ .