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  • Tangents PA and PB are drawn to circle <img alt="\mathrm{(x+2)^2+(y-2)^2=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%28x&plus;2%29%5E2&plus;%28y-2%29%5E2%3D1%

Tangents PA and PB are drawn to circle \mathrm{(x+2)^2+(y-2)^2=1}  from point P which lie on \mathrm{y=x}  line, then the locus of circum centre of \mathrm{\triangle \mathrm{PAB}}  is
 

Option: 1

\mathrm{ y=x+2}


Option: 2

\mathrm{y=x+1}


Option: 3

\mathrm{x^2+y^2+x-y=0}


Option: 4

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


Answers (1)

best_answer

Circum centre \mathrm{(h, k)} of \mathrm{\triangle P A B} is the mid point of line joining (-2,2) and \mathrm{ P(t, t)}.

Posted by

Divya Prakash Singh

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