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A variable circle always touches the line \mathrm{y=x} and passes through the point (0,0). The common chords of above circle and \mathrm{x^2+y^2+6 x+8 y-7=0} will pass through a fixed point, whose coordinates are
 

Option: 1

\mathrm{\left(\frac{1}{2}, \frac{1}{2}\right)}



 


Option: 2

\left(\frac{1}{2},-\frac{1}{2}\right)


Option: 3

\left(-\frac{1}{2},-\frac{1}{2}\right)


Option: 4

\left(-\frac{1}{2}, \frac{1}{2}\right)


Answers (1)

best_answer

Family of circles \mathrm{x^2+y^2+\lambda(x-y)=0}

For common chords \mathrm{S_1-S_2=0 \Rightarrow 6 x+8 y-7-\lambda(x-y)=0}

Point of concurrency is intersection point of lines

\mathrm{6 x+8 y-7=0\: and\: x-y=0}
\mathrm{ \Rightarrow \quad(x, y)=\left(\frac{1}{2}, \frac{1}{2}\right) }

Hence option 1 is correct.

Posted by

Suraj Bhandari

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