Given A = (-1,0) and B = (11,4), how do you show that the equation of the circle with AB as diameter may be written as (x-5)^2 + (y-2)^2 = 40?

Dec 23, 2016

see explanation.

Explanation:

Recall that the center-radius form of the circle equation is in the format:
(x – h)^2 + (y – k)^2 = r^2,
with the center being at the point $\left(h , k\right)$ and the radius being $r$.

Given $A \left(- 1 , 0\right) , \mathmr{and} B \left(11 , 4\right)$,
Let distance between $A \mathmr{and} B$ be $D$.
$\implies D = \sqrt{{\left(11 - \left(- 1\right)\right)}^{2} + {\left(4 - 0\right)}^{2}} = \sqrt{160} = \sqrt{16 \cdot 10} = 4 \sqrt{10}$
Given $D$ is the diameter, $\implies \frac{D}{2} = R$ (radius)
$\implies R = \frac{D}{2} = \frac{4 \sqrt{10}}{2} = 2 \sqrt{10}$
The circle is centered at midpoint of AB :
$\implies$ midpoint of $A B = \left(\frac{- 1 + 11}{2} , \frac{0 + 4}{2}\right) = \left(5 , 2\right)$

So the equation of the circle can be written as :
${\left(x - 5\right)}^{2} + {\left(y - 2\right)}^{2} = {\left(2 \sqrt{10}\right)}^{2}$,
$\implies {\left(x - 5\right)}^{2} + {\left(y - 2\right)}^{2} = 40$ (proved)