What is the equatiob of a circle which has a diameter with endpoints (3,1) and (-7,6)?

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Answer 1 · 2018-08-04T09:33:09

The eqn. of circle is :

#x^2+y^2+4x-7y-15=0#

Here , the circle has a diameter with endpoints

#A(3,1) and B(-7,6)#

So. length of #bar(AB)# is the length of diameter
and midpoint of #bar(AB)# is the center of circle.

#AB=sqrt((3+7)^2+(1-6)^2)=sqrt(100+25)=sqrt125#

#AB=5sqrt5todiameter=5sqrt5#

#=>color(red)(radius =r=(5sqrt5)/2#

Midpoint of #bar(AB)=C# ,then coordinate of center #C# is:

#((3+(-7))/2,(1+6)/2)=(-4/2,7/2)=color(red)(C(-2,7/2)#

#"The eqn. of circle with center C(h,k) and radius r is :"#

#(x-h)^2+(y-k)^2=r^2#

#:.(x-(-2))^2+(y-7/2)^2=((5sqrt5)/2)^2#

#=>x^2+4x+4+y^2-7y+49/4=125/4#

#x^2+4x+4+y^2-7y=125/4-49/4#

#x^2+y^2+4x-7y=76/4-4#

#x^2+y^2+4x-7y=60/4#

#x^2+y^2+4x-7y-15=0#

Answer 2 · 2018-08-04T09:46:48

The eqn. of the circle is :

#x^2+y^2+4x-7y-15=0#

#color(blue)("The eqn. of a circle which has a diameter with endpoints"#

#color(blue)(A(x_1,y_1) and B(x_2,y_2) # is :

#color(red)((x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0#

We have #A(3,1) and B(-7,6)#

So the eqn. of the circle is :

#(x-3)(x-(-7))+(y-1)(y-6)=0#

#=>(x-3)(x+7)+(y-1)(y-6)=0#

#=>x^2+7x-3x-21+y^2-6y-y+6=0#

#=>x^2+y^2+4x-7y-15=0#