Question

Find the angle between the tangents drawn from 3, 2 to the circle x^+y^-6x+4y-2=0​

Answer 1

Step-by-step explanation:

We have,   S=x2+y2−2x+4y−11=0  

And the given point is (1,3)

So S1=(1)2+(3)2−2(1)+4(3)−11=9

and  T=x(1)+y(3)−(x+1)+2(y+3)−11=5y−6

So equation of pair of tangent is given by, SS1=T2

⇒9(x2+y2−2x+4y−11)=(5y−6)2

⇒9x2+9y2−18x+36y−99=25y2−60y+36

⇒9x2−16y2−18x+96y−135=0

Comparing with general second degree equation,

a=9,b=−16,h=0

Thus angle between the tangents =tan−1∣∣∣∣∣∣a+b