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Variable circles which always touches the line \mathrm{x}+2 \mathrm{y}-3=0 \text{ at }(1,1), cuts the circle \mathrm{x^2+y^2+4 x-2 y+1=0}. Then all common chords of circles pass through a fixed point, the co-ordinate of the point is
 

Option: 1

\mathrm{ \left(\frac{1}{2}, \frac{5}{6}\right)}


Option: 2

\mathrm{ \left(\frac{1}{6}, \frac{17}{12}\right)}


Option: 3

\mathrm{ \left(\frac{2}{3}, \frac{1}{5}\right)}


Option: 4

 (3,2)
 


Answers (1)

best_answer

Let equation of variable circles is \mathrm{(x-1)^2+(y-1)^2+\lambda(x+2 y-3)=0} equation of common chord of variable circle and given circle is

\mathrm{S}-\mathrm{S}^{\prime}=0 \Rightarrow \lambda(\mathrm{x}+2 \mathrm{y}-3)+(1-6 \mathrm{x})=0

common chords always pass though point of intersection of \mathrm{1-6 \mathrm{x}=0\ \& \ \mathrm{x}+2 \mathrm{y}-3=0.}

 

Posted by

himanshu.meshram

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