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Equation of the circle having intercept of the line \mathrm{y+3 x=0} by the circle \mathrm{ x^2+y^2+6 x-8 \mathrm{y}=0} as a diameter is
 

Option: 1

\mathrm{x^2+y^2+3 x-9 y=0}

 


Option: 2

\mathrm{x^2+y^2+6 x-8 y+3=0}
 


Option: 3

\mathrm{x^2+y^2+3 x+y=0}
 


Option: 4

\mathrm{x^2+y^2+9 x-7 y=0}


Answers (1)

best_answer

Any circle through the point of intersection of the circle \mathrm{x^2+y^2+6 x-8 y=0} and the line \mathrm{y+3 x=0 \: is\: x^2+y^2+6 x-8 y+\lambda(y+3 x)=0}, for some
\mathrm{\lambda \in R}. Centre of this circle is \mathrm{\left(-3-\frac{3 \lambda}{2}, 4-\frac{\lambda}{2}\right)} Since \mathrm{y+3 x=0} is along a diameter of the required circle,

\mathrm{4-\frac{\lambda}{2}-9-\frac{9 \lambda}{2}=0}    \mathrm{or, \: \: \lambda=-1}

Thus required circle is

\mathrm{x^2+y^2+6 x-8 y-(y+3 x)=0 \: or \: x^2+y^2+3 x-9 y=0}

Hence option 1 is correct.

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