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$left parenthesis 12 square root of 2 right parenthesis$ is a point on circle $x to the power of 2 end exponent plus y to the power of 2 end exponent equals 9$ locate the points on the given circle which are at 2 unit distance from $left parenthesis 12 square root of 2 right parenthesis$

$(- 1,2 \sqrt{2})$
$(2 \sqrt{2}, 1)$
$(\frac{23}{9}, \frac{10 \sqrt{2}}{9})$
$(\frac{23}{9}, \frac{10 \sqrt{2}}{9})$

The correct answer is: $open parentheses fraction numerator 23 over denominator 9 end fraction comma fraction numerator 10 square root of 2 over denominator 9 end fraction close parentheses$

$fraction numerator x subscript 1 end subscript minus 1 over denominator cos invisible function application theta end fraction equals fraction numerator y subscript 1 end subscript minus 2 square root of 2 over denominator sin invisible function application theta end fraction equals 2$
$table row cell x subscript 1 end subscript superscript 2 end superscript plus y subscript 1 end subscript superscript 2 end superscript equals 9 end cell row cell text end text text o end text text n end text text end text text s end text text o end text text l end text text v end text text i end text text n end text text g end text text end text end cell row cell c o s invisible function application theta equals fraction numerator 7 over denominator 9 end fraction text end text text o end text text r end text text end text minus 1 end cell row cell text end text text t end text text h end text text e end text text n end text text end text text g end text text e end text text t end text text end text text t end text text h end text text e end text text end text text p end text text o end text text i end text text n end text text t end text text end text left parenthesis negative 12 square root of 2 right parenthesis comma open parentheses fraction numerator 23 over denominator 9 end fraction comma fraction numerator 10 square root of 2 over denominator 9 end fraction close parentheses text end text text. end text text end text end cell end table$

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