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Consider a family of circles passing through two fixed points A(3,7) and B(6,5). The chords in which the circle x^2+y^2-4 x-6 y-3=0 cuts the members of the family are concurrent. Find the coordinates of the point.

Option: 1

\mathrm{\left(\frac{12}{3}, 6\right)}


Option: 2

\mathrm{\left(\frac{12}{3}, 7\right)}


Option: 3

\mathrm{\left(2, \frac{23}{3}\right)}


Option: 4

\mathrm{\left(7, \frac{23}{3}\right)}


Answers (1)

best_answer

Any circle through A and B is required. Take this as the circle with A B as diameter.

 

\overline{\therefore x-3} \overline{x-6}+\overline{y-7} \overline{y-5}=0$ i.e., $x^2+y^2-9 x-12 y+53=0
Equation to A B  \frac{y-5}{x-6}=\frac{2}{-3}

\begin{array}{r} -3 y+15=2 x-12 \\ \\2 x+3 y-27=0 . \end{array}

The equation to the family is x^2+y^2-9 x-12 y+53+K(2 x+3 y-27)=0

This circle intersects x^2+y^2-4 x-6 y-3=0 along the chord

-5 x-6 y+56+K(2 x+3 y-27)=0

This represents for various values of K (each value of K gives a member of the family) straight lines which pass through the point of intersection of,

-5 x-6 y+56=0
and    2 x+3 y-27=0 i.e., the point \left(2, \frac{23}{3}\right) which is same for all K.

Posted by

Divya Prakash Singh

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