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  • Let <img alt="\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{a} x-\mathrm{b} y+\mathrm{a}}{\mathrm{b} x+\mathrm{c} y+\mathrm{a}}," src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7B%

Let \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{a} x-\mathrm{b} y+\mathrm{a}}{\mathrm{b} x+\mathrm{c} y+\mathrm{a}}, where \mathrm{a}, \mathrm{b}, \mathrm{c} are constants, represent a circle passing through the point (2,5). Then the shortest distance of the point (11,6) from this circle is :

Option: 1

10


Option: 2

8


Option: 3

7


Option: 4

5


Answers (1)

best_answer

Let the equation of the circle be

\mathrm{x^{2}+y^{2}+2 g x+2 f y+k=0}

Passes through \mathrm{\left ( 2,5 \right )}

\mathrm{\Rightarrow 4+25+4 g+10 f+k=0 \Rightarrow 4 g+10 f+k=-2 g}      .......(1)

Differentiating with respect to \mathrm{x},

\mathrm{ 2 x+2 y \frac{d y}{d x}+2 g+2 f \frac{d y}{d x}=0} \\

\mathrm{\Rightarrow \frac{d y}{d x}=\frac{-x-g}{y+f}=\frac{a x-b y+a}{b x+c y+a}}

Comparing \mathrm{b=0, g=1, f=-1, c=1}

\mathrm{K=-29-4+10=-23}

So equation of circle is

\mathrm{x^{2}+y^{2}+2 x-2 y-23=0}

\mathrm{\text { Centre: }(-1,1), \quad \text { Radius }=5 }

Shortest distance of  \mathrm{\left ( 11,6 \right )}  from the circle

\mathrm{13-5=8}

Hence the correct answer is option 2.

 

Posted by

Irshad Anwar

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