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  • The locus of the point from which the length of the tangents to the circle <img alt="\mathrm{x^2+y^2-4=0\: and \: x^2+y^2-8 x+15=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7

The locus of the point from which the length of the tangents to the circle \mathrm{x^2+y^2-4=0\: and \: x^2+y^2-8 x+15=0} are equal is given by the equation
 

Option: 1

\mathrm{x=3}
 


Option: 2

\mathrm{y=5}
 


Option: 3

\mathrm{3 y-7=0}
 


Option: 4

\mathrm{8 x-19=0}


Answers (1)

best_answer

Let the point be (\alpha, \beta)

\mathrm{ \alpha^2+\beta^2-4=\alpha^2+\beta^2-8 \alpha+15 \Rightarrow 8 \alpha=19 . }

Hence the locus of \mathrm{ (\alpha, \beta)\: is \: 8 x=19. }

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