Chapter 1, Problem 104RE

To determine

To calculate:The equation for the given circle with center $\left(2,-5\right)$ and radius $\sqrt{2}$

Answer to Problem 104RE

The equation for the given circle with center $\left(2,-5\right)$ and radius $\sqrt{2}$ is $x^2+y^2-4x+10y+27=0$

Explanation of Solution

Given information:

A circle with center $\left(h,k\right)=\left(2,-5\right)$ and radius $r=\sqrt{2}$

Formula used:

For a given circle with center $\left(h,k\right)$ and radius $r$ the equation is given as:

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

This is referred to as the Standard form for the equation of a given circle.

Moreover, if the center of the circle is nothing but the origin that is $\left(0,0\right)$ then the equation reduces to:

  $x^2+y^2=r^2$

Calculation:

Consider the given circle with center $\left(h,k\right)=\left(2,-5\right)$ and radius $r=\sqrt{2}$

Recall that for a given circle with center $\left(h,k\right)$ and radius $r$ the equation is given as:

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

Therefore, replacing the values we get:

  ${\left(x-2\right)}^2+{\left(y-\left(-5\right)\right)}^2={\left(\sqrt{2}\right)}^2$ $\left(1\right)$

Now using the following formulae:

  $\begin{array}{l}{\left(a+b\right)}^2=a^2+2ab+b^2\\ {\left(a-b\right)}^2=a^2-2ab+b^2\end{array}$

Then, $\left(1\right)$ reduces to:

  $\begin{array}{l}{x}^2+4-4x+y^2+25+10y=2\\ x^2+y^2-4x+10y+29=2\\ x^2+y^2-4x+10y+29-2=0\\ x^2+y^2-4x+10y+27=0\end{array}$

Hence, the equation for the given circle with center $\left(2,-5\right)$ and radius $\sqrt{2}$ is:

  $x^2+y^2-4x+10y+27=0$