# How do you write an equation for a circle whose diameter has endpoints of (-2, 4) and (4, 12)?

Aug 14, 2018

#### Answer:

The eqn . of circle :
${x}^{2} + {y}^{2} - 2 x - 16 y + 40 = 0$

#### Explanation:

Let , $A \left(- 2 , 4\right) \mathmr{and} B \left(4 , 12\right)$ be the endpoints of diameter $\overline{A B}$ and $C$ be the center of given circle.

Let $r$ be the radius of circle.

Here ,

Center $C$ is the midpoint of diameter $\overline{A B}$

So ,

$C \left(\frac{- 2 + 4}{2} , \frac{4 + 12}{2}\right) \equiv \left(1 , 8\right)$

Now ,,

$\text{Radius r=} C A = C B$

$\therefore r = C B = \sqrt{{\left(4 - 1\right)}^{2} + {\left(12 - 8\right)}^{2}} = \sqrt{9 + 16} = 5$

We know that,

$\text{The eqn. of circle with center C(h,k) and radius r }$ is

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

$\therefore {\left(x - 1\right)}^{2} + {\left(y - 8\right)}^{2} = {5}^{2}$

$\therefore {x}^{2} - 2 x + 1 + {y}^{2} - 16 y + 64 = 25$

$\therefore {x}^{2} + {y}^{2} - 2 x - 16 y + 40 = 0$

Aug 14, 2018

#### Answer:

The eqn. of circle is :
$\therefore {x}^{2} + {y}^{2} - 2 x - 16 y + 40 = 0$

#### Explanation:

We know that ,

$\text{If "A(x_1,y_1) and B(x_2,y_2)"are the endpoints of the }$

$\text{diameter of the circle then the eqn. of circle is :}$

$\left(x - {x}_{1}\right) \left(x - {x}_{2}\right) + \left(y - {y}_{1}\right) \left(y - {y}_{2}\right) = 0$

We have diameter $\overline{A B}$ ,with $A \left(- 2 , 4\right) \mathmr{and} B \left(4 , 12\right) .$

So, the eqn. of circle is :

$\left(x - \left(- 2\right)\right) \left(x - 4\right) + \left(y - 4\right) \left(y - 12\right) = 0$

$\left(x + 2\right) \left(x - 4\right) + \left(y - 4\right) \left(y - 12\right) = 0$

$\therefore {x}^{2} - 4 x + 2 x - 8 + {y}^{2} - 12 y - 4 y + 48 = 0$

$\therefore {x}^{2} + {y}^{2} - 2 x - 16 y + 40 = 0$