Question

the polar of the point (2t, t-4) with respect to the circle x^2+y^2-4x-6y+1=0 passes through the point​

Answer 1

Answer:

Step-by-step explanation:

P(2t,t-4)

S= x^2+y^2-4x-6y+1=0

Polar equation s1 = 0

X(2t)+y(t-4)-2(x+2t)-3(y+t-4)+1=0

t(2x+y-7)-(2x+7y-13)=0

No equate above two equations because t should be zero to satisfy the above equation of polar such that

2x+y-7=0

2x+7y-13=0

Solving those two equations y=1 we get and put y=1 then 2x=6

X=3

Therefore (3,1) is the option