Find the equation of the circle which passes through the point (2, -2) and (3, 4) and whose centre lies on the x + y = 2

Answer

A, D

Explanation

The circle with points `P(2t_(1),2//t_(1))` and `Q(2t_(2),2//t_(2))` as diameter is given by
`(x-2t_(1))(x-2t_(2))+(y-(2)/(t_(1)))(y-(2)/(t_(2)))=1" (1)"`
Also, the slope of PQ is given by
`-(1)/(t_(1)t_(2))=1 or t_(1)t_(2)=-1`
Hence, from (1), the circle is
`(x^(2)+y^(2)-8)-2(t_(1)+t_(2))(x-y)=0`
which is of the form `S+lambdaL=0`.
Hence, circles pass through the points of intersection of the circle
`x^(2)+y^(2)-8=0` and the line x = y.
The point of intersection are (2, 2) and `(-2, -2)`.