Question

Find the standard form equation of the circle with endpoints of a diameter at (-1, 5) and (3, -1).

a) (x+1)^2 + (y+2)^2 = 13/2
b) (x-1)^2+ (y-2)^2 = 13
c) (x+1)^2+ (y+2)^2= 13
d) (x-1)^2+ (y+2)^2 = 13
e) (x-1)^2+ (y-2)^2 = 13/2
f) (x+1)^2+ (y-2)^2 = 13/2

Answer 1

==> Standard form equation of a circle is :
\[\Large \color{MidnightBlue}{\bf{(x-a)^2+(y-b)^2=r^2}}\]
Where :
=====
==> (a,b) = Center .
==> ( r ) = Radius .
Now ,You should know that the mid point of the diameter is center of the circle ;
To find the midpoint Use This formula :
\[\Large \color{MidnightBlue}{\bf{MidPoint=\frac{x_{1}+x_{2}}{2}}} >(Fp),\color{MidnightBlue}{\bf\frac{y_{1}+y_{2}}{2}}>(Sp)\]
~substitute and get the center . Note :(Fp)=First Point , (Sp)=Second Point .

==> Let's Move to the radius ,To find out the length of radius by distance formula between any one of the end points and center ,you will need this formula :
\[\Large \color{MidNightBlue}{\bf\sqrt{(x_{1}-(Fp))^2+(y_{1}-(Sp))^2}}\] .
Get The Radius and fill in the Standard form equation of a circle .
-------
Got It ?