1)If 2x² +xy + (  -4)x +6y -5 =0 is the equation of the circle then its radius

 is-                                                                                                                        

2) If the equation of the circle is ax² +(2a -3)y² -4x-1=0 then its centre is-

 

3)Two vertices of an equilateral triangle are (-1,0) $ (1,0) and its third vertex lie above the x-axis.The equation of the circumcircle of the triangle is-

(1) Since coeffecient of xy is 0 so  is 0 so radius g=-(-4)/2=2 f=-3

c= -5/2   now find radius from formula

(2) In eqn of circle coeff. of x^2 and y^2 are same so

2a-3=a or a=3 so centre is (4/3,0) from standard formula

 

(3) thecoordinate of 3rd vertice is (0,y) .since eq. triangle and side length is 2 so

y^2+1=2^2 or y=(3)^(1/2)

now in eq. triangle centroid=circumcentre so from formula (x1+x2+x3)/3 get centroid and then you get (g,f) now in standard circle eqn. put them and then put any of the coordinates of vertices in the eqn. to get c.

ASMIT $ TROY plz answer me. how both of u got diff values of g,f,and c.
According to what u say the vvalue shud be
g=-1
f=3/2
c=-5/2
your ans to my 2nd quest is(4/3,0) but c=(2/3,0)

 

      plz tell me if im wrong, i m in confusion.

3) ans= ^{[ ]}3(x²+y²)-2y -^{[ ]}3=0

 

plz solve q-2)  i m getting (2/3,0)

q3)

 

three vertices of the triangle are (-1,0) , (1,0) and (0,3)

 

Consider another circlw whose diametric points are (-1,0) , (1,0) , it's eqn would be

x² - 1 = 0

 

and eqn of line joining (-1,0) , (1,0) is

x -1 = 0

 

.: any circle passing through the intersection of tis line and the the abv circle is of the form

 

S1+ aL1 = 0

 

x² + y² - 1 + a(x-1) = 0

This passed through 0,root3

 

.: a = 2

 

We have the eqn as

x² + y² + 2x - 3 = 0