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  • If A is a point on the circle <img alt="\mathrm{x^{2}+y^{2}-4 x+6 y-3=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E%7B2%7D&plus;y%5E%7B2%7D-4%20x&plus;6%20y-3%3D

If A is a point on the circle \mathrm{x^{2}+y^{2}-4 x+6 y-3=0}.  which is farthest from the point \mathrm{(7,2)}, then

Option: 1

(2-2 \sqrt{2},-3-2 \sqrt{2})


Option: 2

(2+2 \sqrt{2},-3-2 \sqrt{2})


Option: 3

\quad(2+2 \sqrt{2},-3+2 \sqrt{2})


Option: 4

\quad(2-2 \sqrt{2},-3+2 \sqrt{2})


Answers (1)

best_answer

Given circle is \mathrm{(x-2)^{2}+(y+3)^{2}=16}
slope of PA  Parametric =  \mathrm{\frac{2+3}{7-2}= 1} equation of line PC is


\mathrm{\frac{x-2}{\frac{1}{\sqrt{2}}}=\frac{y+3}{\frac{1}{\sqrt{2}}}=r \Rightarrow(x, y)=\left(2+\frac{r}{\sqrt{2}},-3+\frac{r}{\sqrt{2}}\right)}
Hence the point is \mathrm{(2-2 \sqrt{2},-3-2 \sqrt{2})}
Hence (A) is the correct answer.

 

 

Posted by

Suraj Bhandari

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