Question

Compare the given equation of the circle and then find the value of $a+b+c+d+e$ if the centre of the circle is at origin (2,0) and radius is 6.
given equation : $a{x}^2+b{y}^2+cx+dy+e=0$

Answer 1

Equation of a circle with centre $(h,k)$ and radius r is given by
$(x-h{)}^2+(y-k{)}^2=r^2$
In our case center = (2,0) and radius = 6
so equation of the circle :
$(x-2{)}^2+(y-0{)}^2=6^2$
$x^2+y^2-4x+0y+4=36$
$x^2+y^2-4x+0y-32=0$ is the equation of the circle.
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On comparing with the given equation :
$a=1,b=1,c=-4,d=0,e=-32$
So $a+b+c+d+e$ = $1+1-4+0-32$
= -34
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