# How do you write the equation for a circle having (3,0) and (-2,-4) as ends of a diameter?

Aug 31, 2016

Equation of circle is ${x}^{2} + {y}^{2} - x + 4 y - 6 = 0$

#### Explanation:

As the end points of diameter are $\left(3 , 0\right)$ and $\left(- 2 , - 4\right)$, the center is their midpoint i.e. $\left(\frac{3 - 2}{2} , \frac{0 - 4}{2}\right)$ or $\left(\frac{1}{2} , - 2\right)$.

The distance between $\left(3 , 0\right)$ and $\left(- 2 , - 4\right)$ will be

$\sqrt{{\left(3 - \left(- 2\right)\right)}^{2} + {\left(0 - \left(- 4\right)\right)}^{2}}$

= $\sqrt{25 + 16}$

= $\sqrt{41}$

Hence diameter is $\sqrt{41}$ and radius $\frac{\sqrt{41}}{2}$.

As the center is $\left(\frac{1}{2} , - 2\right)$ and radius is $\frac{\sqrt{41}}{2}$, the equation of circle is

${\left(x - \frac{1}{2}\right)}^{2} + {\left(y + 2\right)}^{2} = {\left(\frac{\sqrt{41}}{2}\right)}^{2}$ or

${x}^{2} - x + \frac{1}{4} + {y}^{2} + 4 y + 4 = \frac{41}{4}$ or

${x}^{2} + {y}^{2} - x + 4 y + 4 + \frac{1}{4} - \frac{41}{4} = 0$ or

${x}^{2} + {y}^{2} - x + 4 y + 4 - 10 = 0$ or

${x}^{2} + {y}^{2} - x + 4 y - 6 = 0$