Question

Find the standard form of the equation of the circle with endpoints of a diameter at the points (3,8) and (-7,2)

Answer 1

$(x+2)^{2}$+$(y-5)^{2}$= 34

The equation of a circle in standard form is

$(x-a)^{2}$+$(y-b)^{2}$=$r^{2}$

where (a , b) are the coordinates of the centre and r is the radius

The centre is at the midpoint of the endpoints and the radius is the distance from the centre to either of the 2 endpoints

Using the midpoint formula

midpoint = [$\frac{1}{2}$(x$x_{1}$+$x_{2}$, $\frac{1}{2}$($y_{1}$+$y_{2}$

where ($x_{1}$,$y_{1}$=(3,8) and ($x_{2}$,$y_{2}$=(-7,2)

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

Calculate r using the distance formula

r = √($x_{2}$-$x_{1}$)²+($y_{2}$-$y_{1}$)²)

= √((3+2)²+(8-5)²) = √(25+9) = √34 ⇒ r² =(√34)² = 34

equation of circle is : (x+2)²+(y-5)² = 34