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The line \mathrm{x+y=5} intersects the circle \mathrm{x^2+y^2-6 x-8 y+21=0} at points A and B, then the locus of the point C such that AC is perpendicular to BC is

 

Option: 1

\mathrm{x^2+y^2-6 x-4 y+11=0}


Option: 2

\mathrm{x^2+y^2-4 x-6 y+11=0}


Option: 3

\mathrm{x^2+y^2+6 x+4 y+11=0}


Option: 4

none of these


Answers (1)

best_answer

Since AB is fixed and AC is perpendicular to BC. So, locus of C is a circle whose diameter is AB. So, family of circles passing through AB is

\mathrm{\begin{aligned} & x^2+y^2-6 x-8 y+21+\lambda(x+y-5)=0 \\ & \Rightarrow x^2+y^2-(6-\lambda) x-(8-\lambda) y+21-5 \lambda=0 \end{aligned}}

So, centre is \mathrm{\left(\frac{6-\lambda}{2}, \frac{8-\lambda}{2}\right)}

Since AB is diameter so centre must lie on AB

\mathrm{\begin{aligned} & \Rightarrow \frac{6-\lambda}{2}+\frac{8-\lambda}{2}=5 \Rightarrow \lambda=2 \\ & \therefore \text { Locus is } x^2+y^2-4 x-6 y+11=0 . \end{aligned}}

 

Posted by

shivangi.shekhar

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