and ( , ) . Equation of the circle in diameter form is ) + = ) + = = . Example . A line cuts a chord of length units on a circle with centre of the circle ( , ) .
Find the equation of the circle in general form. C ( , ) is the centre and cuts a chord AB on the circle. Let M be the midpoint of AB . Then we have AM = BM = .
Now BMC is a right triangle. So, we have CM = ( ) ( ) By Pythogoras theorem BC = BM MC Fig. . C B M - - BC = = radius.
So, the equation of the required circle is ( = = . Example . A circle of radius units touches both the axes. Find the equations of all possible circles formed in the general form.
As the circle touches both the axes, the distance of the centre from both the axes is units, centre can be ( ± ± and hence there are four circles with radius , and the required equations of the four circles are ± ± Example . Find the centre and radius of the circle3 Coefficient of x = Coefficient of y (characteristic (ii) for a second degree equation to represent a circle). That is, and a = . Therefore, the equation of the circle is = = So, centre is − and radius r = .
Example . Find the equation of the circle passing through the points( , ), ( , , and( , ) . Let the general equation of the circle be gx fy + = . ...
( ) It passes through points ( , ),( , and( , ) . Therefore, g f = − , … ( ) g f = − , … ( ) g f = − . … ( ) ( ) – ( ) gives g f = . ...
( ) ( ) – ( ) gives g f = − . ... ( ) ( ) + ( ) gives f = − Fig. .
O C (– ,– ) C ( ,– ) C ( , ) C (– , ) - - Two Dimensional Analytical Geometry - II Substituting f = - in ( ), g =− . Substituting f = - and g =− in ( ) , c = . Therefore, the required equation of the circle is = ⇒ x = . Note Three points on a circle determine the equation to the circle uniquely.
Conversely three equidistant points from a centre point forms a circle.