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The mid-point of the chord AB of the circle x^2+y^2-6 x-4 y+3=0  is the point (1,1). Determine the co-ordinates of the point of intersection of tangents to the circle at its extremities.

Option: 1

\begin{aligned} (h=4, k=2) \\ \end{aligned}


Option: 2

(h=1, k=3) \\


Option: 3

(h=5, k=7)


Option: 4

(h=5, k=7)


Answers (1)

best_answer

If (h,k) be the required point, then AB is chord of contact of (h,k) and its equation is 

also given by T=S_1 ? as (1,1) is its mid-point:

2 x+y=3 \: \: by \: \: T=S_1

x . h+y . k-3(x+h)-2(y+k)+3=0

Or \: \: x(h-3)+y(k-2)=3 h+2 k-3 \: \: as \: \: C.C.

Comparing, we get

2 h-3=1 k-2=33 h+2 k-3

Above will give two equations which when solved give h=1, k=0

Posted by

Suraj Bhandari

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