Let ( a,b be the point on the line x+2y+K=0 We have, equation of circle is x2+y2+8x−6y−24=0 Equation of polar for the given circle is obtained as ‌ax+by+4(x+d)−3(y+b)−24=0 ‌(a+4x+(b−3)y+4a−3−24 Now, comparing this line with 2x−3y+3=0 we get ‌‌
a+4
2
=‌
b−3
−3
=‌
4a−y−24
3
‌−3a−12=2b−6⇒3a+2b=−6 ‌‌ and ‌3a+12=8a−6b−4s ‌⇒5a−6b=60...‌ (ii) ‌ On solving Eqs. (i) and (ii), we get a=3‌ and ‌b=‌
−15
2
Now. (3,‌
15
2
) lies on line x+2y+k=0 3−15+K=0⇒K=12 Required length of tangent from a point ‌(‌
12
4
,‌
12
3
)‌ i.e. ‌(3,4) ‌=√33+42+24−24−24 ‌=√9+16−24=1