TS EAMCET 2017 Code A Solved Paper

Consider the given equation.
x2+y2−6x+4y=12
(x−3)2+(y+2)2=25
(x−3)2+(y+2)2=(5)2
If m is the slope, then the equation of the tangent is given by,
y+2=m(x−3)±5√m2+1
The slope of the tangent is same as the line 4x+3y+5=0
Thus, m=−‌‌

4

3

y+2=−‌‌

4

3

(x−3)±5√(−‌‌

4

3

)2+1
y+2=‌‌

−4

3

(x−3)±5(

5

3

)
3y+6=−4x+12±25
4x+3y=6±25
Therefore, the equation becomes
4x+3y−31=0
4x+3y+19=0