Given the three circles x²+y²-16x+60=0, 3x²+3y²-36x+81=0,x²+y²-16x-12y+84=0,find (i)the point from which the tangents to them are equal in length and (ii)this length.

Can you plz explain it clearly to me to understand?? thanking you Sir..............................................................................................................

The radical axis of two circles has the property that the tangents drawn from a point on it to both the circles will have the same length. Radical axis equation quite naturally is S₁=S₂ i.e. S₁- S₂ =0 which is a straight line. 

 

So, for this problem, find the radical axes of any two pairs of circles and their intersection will give you the point you seek( in this problem though, the point is inside the circle and hence length of tangent is not meaningful)

S1: x^2+y^2-16x+60=0

S2:3x^2+3y^2-36x+81=0 =>x^2+y^2-12x+27=0

S3:x^2+y^2-16x-12y+84=0

 

S1-S2=0

x^2+y^2-16x+60-x^2-y^2+12x-27=0

-4x=-33

x=33/4

 

S2-S3=0

x^2+y^2-12x+27-x^2-y^2+16x+12y-84=0

4x+12y-57=0

 

S3-S1=0

x^2+y^2-16x-12y+84-x^2-y^2+16x-60=0

-12y+24=0

-12y=-24

y=24/12

y=2

 

Therefore, (x,y)=(33/4,2)

FIRST SELECT ANY OF THE EQUATION OF THE CIRLE AND THEN PUT THE VALUE OF (x,y)=(33/4)

 

root S1 =>root x^2+y^2-16x+60

          =>root (33/4)^2+(2)^2-16(33/4)+60

          =>root 1089/16+4-132+60

          =>root 1089/16-68

          =>root 1089-1088/16

          =>root 1/16

          =>1/4