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Consider the following lines

  1. $x-y-1=0$
  2. $x+y-5=0$
  3. $y=4$

The line 1 is the axis of the parabola, the line 2 is the tangent at the vertex to the same parabola, and the line 3 is another tangent to the same parabola at some point $P$.

Now let a circle $C$ circumscribe the triangle formed by tangent and normal at the point $P$ and the axis of the parabola.

Then how can I find the equation of the circle?

I found that the vertex of this parabola is $(3,2)$. Don't know how to proceed further.

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    $\begingroup$ It's better to close in the opposite direction: this copy has a reasonable answer. $\endgroup$ – user147263 Oct 20 '15 at 13:19
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HINT

I would say, help subtangent of parabola:

enter image description here

axis: $ x-y-1=0,\quad t_V:\,x+y-5=0, \quad t_P:\, y-4=0$

Theorem: Vertex bisects subtangent of parabola. -->

$P_1 = t_P \times \,$axis, $\quad\overline{P_1V}=\overline{VP_2}\Rightarrow P_2$

Perpendicular from P_2 to axis $\times\, t_P = P, \quad$ perpendicular from P to $t_P \times\,$axis $ = P_3$

$P_2$ = mid searched circle, = focus of parabola.

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  • $\begingroup$ Thanks brother for your help.... $\endgroup$ – Aditya Oct 19 '15 at 18:10

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