Chapter 1.4, Problem 62AYU

To determine

To explain: why the given center and radius values are incorrect.

Answer to Problem 62AYU

The given center and radius values are incorrect.

The correct center is $\left(-3,2\right)$ .

Explanation of Solution

Given:

  ${\left(x+3\right)}^2+{\left(y-2\right)}^2=16$

Center = $\left(3,-2\right)$

Radius = $4$

First compare the equation ${\left(x+3\right)}^2+{\left(y-2\right)}^2=16$ to the standard form of the equation of a circle with radius $r$ and center $\left(h,k\right)$ ;

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

The comparison yields information about the circle.

  $h=-3,k=2$ and $r=4$ .

Therefore, the center is $\left(3,-2\right)$ and radius is $4$ units.

Hence, the center $\left(3,-2\right)$ of equation ${\left(x+3\right)}^2+{\left(y-2\right)}^2=16$ is incorrect. The correct center is $\left(-3,2\right)$ .

Conclusion:

Hence,the given center and radius values are incorrect.

The correct center is $\left(-3,2\right)$ .