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  • The circle passing through the intersection of the circles, <img alt="\mathrm{x^2+y^2-6 x=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E2-6%20x%3D0%7D

The circle passing through the intersection of the circles, \mathrm{x^2+y^2-6 x=0} and \mathrm{x^2+y^2-4 xy=0} having its centre on the line,\mathrm{2x-3y+12=0},also passes through the point 

Option: 1

(-1,3)


Option: 2

(-3,6)


Option: 3

(-3,1)


Option: 4

(1,-3)


Answers (1)

best_answer

Let the family of circles be

\mathrm{\begin{aligned} & S_1+\lambda S_2=0 \\ & \Rightarrow x^2+y^2-6 x+\lambda\left(x^2+y^2-4 y\right)=0 \\ & \text { Centre }=\left(\frac{3}{1+\lambda}, \frac{2 \lambda}{1+\lambda}\right) \end{aligned}}-------(i)

Centre lies on \mathrm{2 x-3 y+12=0}

\mathrm{\begin{aligned} & \text { Then } \frac{6}{\lambda+1}-\frac{6 \lambda}{1+\lambda}+12=0 \\ & \Rightarrow \lambda=-3 \end{aligned}}

Equation of circle (i) is

\mathrm{x^2+y^2+3 x-6 y=0}

point (–3, 6) satisfy above circle.

 

Posted by

Ritika Harsh

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