Prove that the point (-1,-2) lies on the circle x^2+y^2-x-y-8=0.find the co-ordinates of the other extremity of the diameter through (-1,-2)?

1 Answer
F. Javier B.
Feb 12, 2018

See explanation below

Explanation:

The point #P(-1,-2)# lies on circunference (not circle. Circle is a surface not a line) due to

#1^2+2^2+1+4-8=0#. So #P# verifies the equation of circunference and hence P belongs to circunference.

Now, let complete the squares in the equation

#x^2+y^2-x-y-8# = #x^2-x+1/4+y^2-y+1/4-8-1/2# Thus we have #(x-1/2)^2+(y-1/2)^2= sqrt(17/2)^2#. The circunference has his center at #(1/2,1/2)# and his radius is #17/2#

So, we have to obtain the straight line equation passing by #(1/2,1/2)# and #P(-1,-2)#. The interception between this line and circunference will be the coordinates of other extremity diameter.

Let see: The parametric equations of this straight line are:

#x=-1+3/2lambda#
#y=-2+5/2lambda#

Substituting in circunference equation, we have (calculating and ordering terms)

#lambda(34/4lambda-17)=0# with two solutions for #lambda#

#lambda=0# wich is the point P
#lambda=2# wich is the other asked extremity diameter #(2,3)#

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