How do you write an equation for a circle with diameter has endpoints(-3,4) and (4,-3)?

1 Answer
May 23, 2016

Answer:

The equation of the circle is:

#color(brown)(=>49/2=(x_i-1/2)^2+(y_i-1/2)^2)#

Explanation:

The centre will be at the middle of these points so it is the mean value.

Let point 1 be #P_1->(x_1,y_1)->(-3,4)#
Let point 2 be #P_2->(x_2,y_2)->(4,-3)#
Let the centre be #P_c->(x_x,y_c)#
Let the radius be #r#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the centre")#

Mean for x#-> (4+(-3))/2= 1/2 ->x_c#

Mean for y#->(4+(-3))/2=1/2->y_c#

#color(blue)("Centre "-> (x_c,y_c)->(1/2,1/2))#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the magnitude of the radius")#
Using Pythagoras

I chose #(x_1,y_1)# to determine the length of #r#

#r^2=(x_c-x_1)^2+(y_c-y_1)^2#

#r^2=[1/2-(-3)]^2+[1/2-4]^2#

#r=sqrt(49/4+49/4) = sqrt(49/2)#

#r=7/sqrt(2)#

Multiply by 1 but in the form of #1=sqrt(2)/sqrt(2)#

#color(blue)(r=(7sqrt(2))/2" "->" " r^2=49/2)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Putting it all together")#

Let any #x# be #x_i#
Let any #y# be #y_i#

The equation of this circle is

#r^2=(x_i-x_x)^2+(y_i-y_c)^2#

#color(blue)(=>49/2=(x_i-1/2)^2+(y_i-1/2)^2)#