Math, asked by mohimohith070, 3 days ago

Position of point (4,2) w.r.to 2x²+2y²-5x-4y-3=0 is​

Answers

Answered by itsRakesh
2

Answer:

2x^{2}+2y^{2}-5x-4y-3=0 is a circle, and (4,2) is a point which lies

outside the circle

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