Let P(h, k) be a point on the circle 15x2+15y2−48x+64y=0 Then the lengths of the tangents from P(h, k) to 5x2+5y2−24x+32y+75=0 and 5x2+5y2−48x+64y+300=0 are
PT1=√h2+k2−‌‌
24
5
h+‌‌
32
5
k+15 and PT2=√h2+k2−‌‌
48
5
h+‌‌
64
5
k+60 or PT1=√‌‌
48
15
h−‌‌
64
15
k−‌‌
24
5
h+‌‌
32
5
k+15=√‌‌
32
15
k−‌‌
24
15
h+15
(Since (h, k) lies on 15x2−15y2−48x+64y=0 ∴h2+k2−‌‌