TS EAMCET 5 May 2018 Shift 2 Solved Paper

Consider the equation of circle.
x2+y2−6x−4y+4=0
Its centre is (3,2) and radius is r=3
Let P(h,k) be the point on the line 4x−3y=6 from which the tangent are drawn to circle as shown in figure below.Let the angle between the tangents be 2α .
Therefore,
2α=tan−1

24

7

tan‌2‌α=

24

7

2‌tan‌α

1−tan2α

=

24

7

7‌tan‌α=12−12tan2α
Solve further,
12tan2α+7‌tan‌α−12=0
(4‌tan‌α−3)(3‌tan‌α+4)=0
Therefore,
tan‌α=

3

4

Consider the figure shown below,sin‌α=

3

√(h−3)2+(k−2)2

3

5

=

3

√(h−3)2+(k−2)2

(h−3)2+(k−2)2=25
h2+k2−6h−4k=12 …… (I)
Now (h,k) lies on 4x−3y=6
Hence,
4h−3k=6…… (II)
Solve equation (I) and (II).
When k=6 then h=6
And
When k=−2 then h=0.