EAMCET Engineering 2010 Solved Paper

Let P(x1,y1) be the point from which the tangents are drawn to the circles
‌S1≡x2+y2−8x+40=0
‌S2≡5x2+5y2−25x+80=0
‌S3≡x2+y2−8x+16y+160=0
Since, the length of the tangent from P to the circle S1,S2,S3 are equal
∴‌√S1=√S2=√S3
⇒S1=S2=S3
x12+y12−8x1+40=5x12+5x22−25x1+80
=x12+y2−8x1+16y1+160 . . . (i)
Taking first and third part of above relation (i), we get
−40+16y1+160=0
16y1+120=0
y1=−‌

120

16

⇒−‌

15

2

Taking first and second part of the relation (i).
−3x1+24=0
x1=‌

24

3

⇒x1=8
Hence, the point P is (8,‌

−15

2

).