# 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?

Jan 27, 2017

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

${\left(x - a\right)}^{2} + {\left(y - a + 2\right)}^{2} = {\left(A C\right)}^{2} = {\left(a - 0\right)}^{2} + {\left(\left(a - 2\right) - 1\right)}^{2} = 2 {a}^{2} - 6 a + 9$,

giving

${x}^{2} + {y}^{2} - 2 a x - 2 \left(a - 2\right) y + 2 a - 5 = 0$

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]}