# How do you find the equation of the circle given diameter at (4, -3) and (-2, 5)?

Jun 14, 2016

${\left(x - 1\right)}^{2} + {\left(y - 1\right)}^{2} = 25$

#### Explanation:

If the diameter is color(brown)(""(4,-3)) to color(orange)(""(-2,5))
then center is at
$\textcolor{w h i t e}{\text{XXX}} \left(\frac{\textcolor{b r o w n}{4} + \left(\textcolor{\mathmr{and} a n \ge}{- 2}\right)}{2} , \frac{\textcolor{b r o w n}{- 3} + \textcolor{\mathmr{and} a n \ge}{5}}{2}\right) = \left(\textcolor{red}{1} , \textcolor{b l u e}{1}\right)$
and
$\textcolor{w h i t e}{\text{XXX}} \sqrt{{\left(\textcolor{b r o w n}{4} - \textcolor{red}{1}\right)}^{2} + {\left(\textcolor{b r o w n}{- 3} - \textcolor{b l u e}{1}\right)}^{2}} = \sqrt{9 + 16} = \textcolor{g r e e n}{5}$

The general equation of a circle
with center $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$ and radius $\textcolor{g r e e n}{r}$ is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \textcolor{red}{a}\right)}^{2} + {\left(y - \textcolor{b l u e}{b}\right)}^{2} = {\textcolor{g r e e n}{r}}^{2}$

In this case we have
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \textcolor{red}{1}\right)}^{2} + {\left(y - \textcolor{b l u e}{1}\right)}^{2} = {\textcolor{g r e e n}{5}}^{2}$

Jun 14, 2016

${x}^{2} + {y}^{2} - 2 x - 2 y - 23 = 0$,

#### Explanation:

Formula for the eqn. of a circle having $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right)$ as diametrically opposite points (pts.) : $\left(x - {x}_{1}\right) \left(x - {x}_{2}\right) + \left(y - {y}_{1}\right) \left(y - {y}_{2}\right) = 0.$

Accordingly, the reqd. eqn. is $\left(x - 4\right) \left(x + 2\right) + \left(y + 3\right) \left(y - 5\right) = 0$,
i.e., ${x}^{2} - 4 x + 2 x - 8 + {y}^{2} + 3 y - 5 y - 15 = 0$,
or, ${x}^{2} + {y}^{2} - 2 x - 2 y - 23 = 0$,

Aliter :-

Extremities of a diameter are $\left(4 , - 3\right) , \left(- 2 , 5\right)$

Centre, being midpoint of diameter, we get the centre $\left(\frac{4 - 2}{2} , \frac{- 3 + 5}{2}\right) = \left(1 , 1\right) .$

Radius = Distance between the Centre & any end point of given diameter $= \sqrt{{\left(4 - 1\right)}^{2} + {\left(- 3 - 1\right)}^{2}} = 5.$

Hence the eqn. $: {\left(x - 1\right)}^{2} + {\left(y - 1\right)}^{2} = {5}^{2} ,$ or, ${x}^{2} + {y}^{2} - 2 x - 2 y - 23 = 0 ,$ as before!