Chapter 1.2, Problem 69E

To determine

ToWrite:Standard form of the equation of the circle with center $\left(-4,5\right)$ and radius $2$

Answer to Problem 69E

The standard form of the equation of a circle with center $\left(-4,5\right)$ and radius $2$ is ${\left(x+4\right)}^2+{\left(y-5\right)}^2=4$

Explanation of Solution

Given:

A circle with center $\left(-4,5\right)$ and radius $2$

Concept Used:

The standard form of the equation of a circle with center $\left(h,k\right)$ and radius $r$ is:

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

Calculation:

Given a circle with center $\left(-4,5\right)$ and radius $2$

Since, the standard form of the equation of a circle with center $\left(h,k\right)$ and radius $r$ is:

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

Substituting the given values of center and radius, we have

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

i.e., ${\left(x+4\right)}^2+{\left(y-5\right)}^2=2^2$

i.e., ${\left(x+4\right)}^2+{\left(y-5\right)}^2=4$

Therefore, thestandard form of the equation of a circle with center $\left(-4,5\right)$ and radius $2$ $7$ is ${\left(x+4\right)}^2+{\left(y-5\right)}^2=4$

Conclusion:

The standard form of the equation of a circle with center $\left(-4,5\right)$ and radius $2$ is ${\left(x+4\right)}^2+{\left(y-5\right)}^2=4$