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  • The limiting points of the coaxial system containing the two circles <img alt="\mathrm{x^2+y^2+2 x-2 y+2=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5

The limiting points of the coaxial system containing the two circles \mathrm{x^2+y^2+2 x-2 y+2=0} and \mathrm{25\left(x^2+y^2\right)-10 x-80 y+65=0}

Option: 1

(1,-1),(-5,-40)


Option: 2

(1,-1),(-1 / 5,-8 / 5)


Option: 3

(-1,1),(1 / 5,8 / 5)


Option: 4

(-1,1),(-1 / 5,-8 / 5)


Answers (1)

best_answer

:The equation of the radical axis of the given circles is 4x + 2y – 1 = 0. Therefore, the equation of family of coaxial circles is

\mathrm{\begin{gathered} x^2+y^2+2 x-2 y+2+\lambda(4 x+2 y-1)=0 \\ \Rightarrow x^2+y^2+2 x(1+2 \lambda)+2 y(\lambda-1)+2-\lambda=0 \end{gathered}}

The coordinates of the centre and radius of this circle are

\mathrm{(-(2 \lambda+1),-(\lambda-1)) \text { and } \sqrt{(2 \lambda+1)^2+(\lambda-1)^2-2+\lambda}}

For limitting points, we must have

Radius =0

\mathrm{\begin{aligned} & \Rightarrow(2 \lambda+1)^2+(\lambda-1)^2+\lambda-2=0 \\ & \Rightarrow \lambda=0,-3 / 5 \end{aligned}}

Hence , the limiting points are (-1,1) and (1/5,8/5).

 

Posted by

Shailly goel

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