Chapter 9, Problem 4CR

To determine

To find equation of a circle with Center $\left(4,-3\right)$ and radius $5$ .

Answer to Problem 4CR

The answer is $b.$

  $x^2+y^2-8x+6y=0$ .

Explanation of Solution

Given:

Center $=$$\left(4,-3\right)$

Radius $=$$5$

Formula Used:

Circle equation is given as ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ ; where $\left(h,k\right)$ is the Center and $r$ is the radius.

Calculation:

Given:

Center $=$$\left(4,-3\right)$

Radius $=$$5$

Using the above formula and putting the above values in the equation of the Circle, we have:

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

  ${\left(x-4\right)}^2+{\left(y+3\right)}^2=5^2$

Solving the above equation, we have:

  $\begin{array}{l}{\left(x-4\right)}^2+{\left(y+3\right)}^2=5^2\\ x^2+16-8x+y^2+9+6y=25\\ x^2-8x+y^2+6y+25-25=0\\ x^2+y^2-8x+6y=0\end{array}$

Equation of the Circle is $x^2+y^2-8x+6y=0$ .

Thus, the answer is $b.$

  $x^2+y^2-8x+6y=0$

Conclusion:

The answer is $b.$

  $x^2+y^2-8x+6y=0$ .