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Equation to the circles pass through the point (2,3) and cut off equal chords of length 6 units along the lines \mathrm{y}-\mathrm{x}-1=0 and \mathrm{x}+\mathrm{y}-5=0 is

Option: 1

\mathrm{(x-3)^{2}+(y-2 \sqrt{3})^{2}=(3 \sqrt{2})^{2}}


Option: 2

\mathrm{(x-2)^{2}+(y-3-3 \sqrt{2})^{2}=(3 \sqrt{2})^{2}}


Option: 3

(\mathrm{x}-2 \sqrt{3})^{2}+(\mathrm{y}-3)^{2}=(3 \sqrt{2})^{2}


Option: 4

none of these


Answers (1)

best_answer

the given two lines pass through the point (2,3) and are inclined at 45^{\circ} and 135^{\circ} to the \mathrm{x}-axis. The co-ordinates of other ends can easily be calculated as

 (2+3 \sqrt{2}, 3+3 \sqrt{2}) and
(2-3 \sqrt{2}, 3+3 \sqrt{2}) . There is a symmetry about the line \mathrm{x}=2 and therefore the centre of the circle lie on \mathrm{x}=2.
As the chords subtend a right angles at the centres

2 \mathrm{r}^{2}=6^{2} \Rightarrow \mathrm{r}=3 \sqrt{2}  centre is
(2,3+3 \sqrt{2})

Posted by

sudhir.kumar

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