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Given the three circles x 2 +y 2 -16x+60=0, 3x 2 +3y 2 -36x+81=0,x 2 +y 2 -16x-12y+84=0,find (i)the point from which the tangents to them are equal in length and (ii)this length.

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.

Grade:12

5 Answers

Nishant Vora IIT Patna
askIITians Faculty 2467 Points
7 years ago
let that point be (h.k)

Now we know that Length of tangent is given by root(S1)

So length of tangent to circle 1 is root(h2+k2-16h+60)

Similarily for other two and then equate all the lengths

Hansraj Gyanendra Singh Rajawat
35 Points
7 years ago
Can you plz explain it clearly to me to understand?? thanking you Sir..............................................................................................................
mycroft holmes
272 Points
7 years ago
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)
dhruv
12 Points
6 years ago
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)
dhruv
12 Points
6 years ago
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

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