Q 28. - Precalculus Enhanced with Graphing Utilities

Question

In Problem, (a) find the center $\left(h,k\right)$ and radius $r$ of each circle; (b) graph each circle; (c) find the intercepts, if any.

$x^2+y^2-6x+2y+9=0$

Step-by-Step Solution

Answer

(a) Center is $\left(3,-1\right)$ and radius is $1$ unit.

(b) Graph is as follows;

(c) $x-$intercept is $\left(3,0\right)$ and no $y-$intercept.

Step 1. Given information

An equation of a circle is $x^2+y^2-6x+2y+9=0$.

Step 2. Center and radius of a circle.

Consider the equation of a circle: $x^2+y^2-6x+2y+9=0$

$\begin{array}{rcl}\left(x^2-2·3·x+3^2\right)-3^2+\left(y^2+2·1·y+1^2\right)-1^2+9& =& 0\\ (x-3{)}^2-9+(y+1{)}^2-1+9& =& 0\\ (x-3{)}^2+(y+1{)}^2& =& 1\end{array}$

The above equation is standard form of the circle with radius $1$ and center $(3,-1)$.

Step 3. Graph of the circle.

Graph of a circle is as follows:

Step 4. Intercepts of a circle.

To find the $x-$intercepts, substitute $y=0$ and solve for $x$.

$\begin{array}{rcl}(x-3{)}^2+(0+1{)}^2& =& 1^2\\ (x-3{)}^2+1& =& 1\\ x-3& =& 0\\ x& =& 3\end{array}$

Therefore, the $x-$intercept is $\left(3,0\right)$.

To find the $y-$intercepts, substitute $x=0$ and solve for $y$.

$\begin{array}{rcl}(0-3{)}^2+(y+1{)}^2& =& 1^2\\ 9+(y+1{)}^2& =& 1\\ (y+1{)}^2& =& -8\end{array}$

Note that square of a number is never negative. Therefore, this equation has no solution and there are no $y$ intercepts.