How do you find an equation of the circle that satisfies the given conditions: endpoints of a diameter are P(−1, 2) and Q(7, 8)?

Apr 16, 2018

$\textcolor{b l u e}{{\left(x - 3\right)}^{2} + {\left(y - 5\right)}^{2} = 25}$

Explanation:

The coordinates of the centre of the circle will be the coordinates of the midpoint of the diameter.

$\left(\frac{- 1 + 7}{2} , \frac{2 + 8}{2}\right) = \left(3 , 5\right)$

We next find the length of the diameter using the distance formula:

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

$d = \sqrt{{\left(7 - \left(- 1\right)\right)}^{2} + {\left(8 - 2\right)}^{2}} = \sqrt{100} = 10$

This is the length of the diameter, so radius is:

$r = \frac{10}{2} = 5$

The equation of a circle is given as:

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

Where $h \mathmr{and} k$ are the $x \mathmr{and} y$ coordinates of the centre respectively.

$\therefore$

${\left(x - 3\right)}^{2} + {\left(y - 5\right)}^{2} = 25$

Apr 16, 2018

${\left(x - 3\right)}^{2} + {\left(y - 5\right)}^{2} = 25$

Explanation:

Equation of a circle is of the form: ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$
Where (a,b) represents the co ordinates of the circle's centre
Equation to find the coordinates of the centre of a line segment is
= $\left(\frac{x 1 + x 2}{2}\right) , \left(\frac{y 1 + y 2}{2}\right)$
Hence,
( 0.5 ( -1 + 7 ) ) , ( 0.5 ( 2 + 8 ) )
= ( 3 , 5 )
For the distance between two points , $\sqrt{{\left(x 2 - x 1\right)}^{2} + {\left(y 2 - y 1\right)}^{2}}$
Hence, sqrt( ( 7 - ( -1 ) )^2 + ( 8 - 2 )^2
= 10
Diameter of circle = 10
= ${\left(x - 3\right)}^{2} + {\left(y - 5\right)}^{2} = {5}^{2}$
= ( x - 3 )^2 + ( y- 5 )^2 = 25