What is the standard form of the equation of a circle passing through A(0,1) , B(3,-2) and has its centre lying on the line y=x-2?

1 Answer
Jan 27, 2017

Answer:

A family of circles #f(x, y; a)=x^2+y^2-2ax-2(a-2)y+2a-5=0#, where a is the parameter for the family, at your choice. See the graph for two members a = 0 and a = 2.

Explanation:

The slope of the given line is 1 and the slope of AB is -1.

It follows that the given line should pass through the midpoint of

M(3/2, -1/2) of AB..

And so, any other point C(a, b) on the given line, with #b = a-2#,

could be the center of the circle.

The equation to this family of circles is

#(x-a)^2+(y-a+2)^2= (AC)^2=(a-0)^2+((a-2)-1)^2=2a^2-6a+9#,

giving

#x^2+y^2-2ax-2(a-2)y+2a-5=0#

graph{(x+y-1)(x-y-2)(x^2+y^2-4x-1)(x^2+y^2+4y-5)=0x^2 [-12, 12, -6, 6]}