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The line y=x touches a circle at point P in the first quadrant such that the distance of P from the origin is 5 \sqrt{2}. If the length of the portion intercepted by this circle on the y-axis is 2 c, then the equation of the circle is (x- 5)^2+(y-5)^2+\left(k_1-k_2 \sqrt{c^2+50}\right)(y-x)=0 where k_1-k_2 is equal to:

 

Option: 1

8


Option: 2

12


Option: 3

10


Option: 4

None of these.


Answers (1)

best_answer

Equation of the family of circles touching y=x at $(5,5)$ can be written as (x-5)^2+(y-5)^2+\lambda(y-x)=0
Put x=0, we get
\begin{aligned} & y^2-10 y+\lambda y+50=0 \quad \Rightarrow \quad \lambda^2-20 \lambda-\left(4 c^2+100\right)=0 \\ & \Rightarrow \quad \lambda=\frac{20 \pm \sqrt{400+4\left(4 c^2+100\right)}}{2} \quad \Rightarrow \quad \lambda=10 \pm 2 \sqrt{c^2+50} \end{aligned}
Since \mathrm{y}_1, \mathrm{y}_2 are positive
\Rightarrow \quad \lambda=10-2 \sqrt{c^2+50}
\therefore \quad Equation of the required circle is
(x-5)^2+(y-5)^2+\lambda(y-x)=0

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Rishabh

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