If the end points of a diameter of circle are `A(x_1,y_1)` and `B(x_2,y_2)` then show that equation of circle will be `(x-x_1)·(x-x_2)+(y-y_1)(y-y_2) = 0`

Centre of the circle (x - x_(1)) (x-x_(2)) + (y-y_(1)) (y- y_(2)) = 0 is

Show that equations of a circle with end points of diameter (x_(1),y_(1))and(x_(2),y_(2)) is same as the equation of circle with end points of diameter (x_(1),y_(2))and(x_(2),y_(1)) . Give reason.

Find the equation of a circle, the co-ordinates of the ends of whose diameter are (-1,-3) and (2,5) Now the equation of circle (x-x_(1))(x-x_(2))+(y-y_(1))(y-y_(2))=0 rArr(x+1)(x-2)+(y+3)(y-5)=0 rArr x^(2)-x-2+ y^(2)-2y-15=0 rArr x^(2)+y^(2)-x-2y-17=0 We is the required equation.

The equation of a circle whose end the points of a diameter are (x_(1), y_(1)) and (x_(2),y_(2)) is

A(x_(1) y_(1)) and B(x_(2), y_(2)) are two given points on a circle. If the chord AB substends an angle theta at a point p on its circumferene, show that the equation of the circle is (x-x_(1)) (x-x_(2))+(y-y_(1))(y-y_(2))=+- cos theta [ (x-x_(1))(y-y_(2))-(x-x_(2))(y-y_(1))]

A: The equation of the circle (2,-3), (-3,2) as ends of a diameter is x^(2)+y^(2)+x+y-12=0 R: The equation of the circle having the line segment joining A(x_(1),y_(1)) and B(x_(2),y_(2)) as diameter (x-x_(1)) (x-x_(2))+(y-y_(1)) (y-y_(2))=0

If (x_1,y_1) and (x_2,y_2) are the end points of a focal chord of the parabola y^2=5x , then 4x_1 x_2+y_1y_2=