Chapter 1, Problem 11CT

To determine

The general form of the circle with center $\left(4,-3\right)$ and radius $5$ .

Answer to Problem 11CT

Solution:

The general form of the circle is $x^2+y^2-8x+6y=0$ .

Explanation of Solution

Given information:

The center of the circle is $\left(4,-3\right)$, and its radius $5$ .

The standard form of an equation of a circle with radius $r$ and center $\left(h,k\right)$ is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$, and by simplifying the standard form of an equation, it gives general form of the equation of a circle.

By comparing $\left(4,-3\right)$ with $\left(h,k\right)$, it gives $h=4$ and $k=-3$ .

By substituting radius of circle and the values $h=4$ and $k=-3$ in ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$, it gives

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

By simplifying,

  $\Rightarrow {\left(x-4\right)}^2+{\left(y+3\right)}^2=25$ .

  $\Rightarrow \left(x^2-8x+16\right)+\left(y^2+6y+9\right)=25$ ,

  $\Rightarrow x^2+y^2-8x+6y+25=25$ ,

  $\Rightarrow x^2+y^2-8x+6y=0$ .

Therefore, the general form of the equation of the circle is $x^2+y^2-8x+6y=0$ .