IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

deep01

If the equation of the incircle of an equilateral triangle is x²+y²+4x-6y+4=0,then the equation of it's circumcircle is??

pls also explain ur answer

ankurgupta91

as in a equilateral triangle the incentre nad circumcente coincides
thus circumcentre O (-2,3) frm the equation of incircle.
nw the radius of incircle is 3
and in equilateral triangle
radius of circumcircle is two times the radius of incircle
radius of circumcircle is 6
the equaiton of circumcirle is
x^2 + y^2 +4x -6y -23 =0
thats the rght answer
thanks.....

siddharthsaxena

i am in huury, i cn tell u d method. first find the radius of this circle.incentre is equidistant frm d sides. hence u will gt 3 equations for 3 sides in x and y. we know that coordinates fr incentre is x= ax1 + bx2 +cx3 / (a+b+c) ; y= ay1 + by2+ cy3 /(a+b+c) where a,b,c are lengths of BC AC and AB resp. u will get 3 more eqns for 3 sides. solve them to get the eqns of sides and find their pt of intersection.it will be the vertices. now the circumcentre is equidistant from 3 vertices and hence u will get its radius and centre and hence d eqn

amar.gupta

very good work ankur. but i think you have done a small calculation mistake.

as radius of incircle = sqrt( 4+9-4) =3 not (9)

so radius of circumcircle will be = 2*3 =6

otherwise logic and procedure is fine .

so good work again.

shan

@ above person:

He has used radius of circumcircle as 6 only 'na. So its right completely.

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