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=