AP EAPCET 18 May 2024 Shift 1 Solved Paper

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