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  • Position of the point <img alt="\mathrm{P\left(1, \frac{\sqrt{11}+6}{6}\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BP%5Cleft%281%2C%20%5Cfrac%7B%5Csqrt%7B11%7D&plu

Position of the point \mathrm{P\left(1, \frac{\sqrt{11}+6}{6}\right)} with respect to circle \mathrm{3 x^2+3 y^2-5 x-6 y+4=0}.

Option: 1

Lies on circle.


Option: 2

Lies outside circle.


Option: 3

Lies inside circle.


Option: 4

None of these.


Answers (1)

best_answer

Dividing by 3 , we get circle as \mathrm{C\left(\frac{5}{6}, 1\right), r=\frac{1}{6} \sqrt{13}} and given point is P.


P will be outside, on or inside the circle according as \mathrm{C P^2>=r^2 ~or ~C P^2<r^2. ~But ~C P^2=\frac{12}{36}<\frac{13}{36}=r^2}


Since \mathrm{C P^2<r^2}, the point P lies inside the circle.

Posted by

Anam Khan

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