Chapter 9, Problem 10RE

To determine

To write: The equation of the circle in standard form.

Answer to Problem 10RE

The standard equation of the circle is ${\left(x+4\right)}^2+{\left(y-3\right)}^2={\left(\frac{7}{2}\right)}^2$

Explanation of Solution

Given information:

Given equation of circle is $4x^2+4y^2+32x-24y+51=0$

Consider the given equation of the circle as,

  $4x^2+4y^2+32x-24y+51=0$  ...... (1)

Divide equation (1) by 4,

  $\begin{array}{c}{x}^2+y^2+8x-6y+\frac{51}{4}=0\\ x^2+8x+16-16+y^2-6y+9-9+\frac{51}{4}=0\\ {\left(x+4\right)}^2+{\left(y-3\right)}^2+\frac{51}{4}-16-9=0\\ {\left(x+4\right)}^2+{\left(y-3\right)}^2=\frac{49}{4}\\ {\left(x+4\right)}^2+{\left(y-3\right)}^2={\left(\frac{7}{2}\right)}^2\end{array}$

This implies that the centre of the circle is $\left(-4,3\right)$ and radius is $\frac{7}{2}$ .

Therefore, the standard equation of the circle is ${\left(x+4\right)}^2+{\left(y-3\right)}^2={\left(\frac{7}{2}\right)}^2$