This downside appeared on the 2022 JEE Superior Paper 1 Arithmetic part as Query 8. Due to Rahul and Battu for the suggestion! The JEE Superior is an especially tough examination for admission into India’s prestigious IIT colleges. This is among the simpler questions of the paper. I give credit score to the Unacademy Atoms video concerning the paper which helped me perceive the best way to resolve the issue.

Let *ABC* be the triangle with *AB* = 1, *AC* = 3, and ∠*BAC* = π/2. If a circle of radius *r* > 0 touches the edges *AB*, *AC* and in addition touches internally the circumcircle of the triangle *ABC*, then the worth of *r* is ___.

As typical, watch the video for an answer.

**JEE 2022 Triangle Circumcircle And Inscribed Circle**

Or hold studying.

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.**Reply To JEE 2022 Triangle Circumcircle And Inscribed Circle**

(Just about all posts are transcribed rapidly after I make the movies for them–please let me know if there are any typos/errors and I’ll right them, thanks).

Since *BAC* = π/2, angle *A* is a proper angle and *BC* is its hypotenuse equal to √(3^{2} + 1^{2}) = √10. Then angle *A* is 90 levels, so it is going to subtend a 180 diploma arc of the triangle’s circumcircle. This implies *BC* is a diameter, the circumcircle’s radius is half its size (0.5√2), and the circumcircle’s heart *M* is the midpoint of the hypotenuse–which is midway horizontally between *A* and *C* (distance 0.5) and midway vertically between *A* and *C* (distance 1.5).

The circle *D* is constructed to the touch sides *AB*, *AC* and is internally tangent at *T* to the circumcircle. As a result of *M* and *D* are tangent circles, their facilities and the tangent level are collinear. Since *MT* is a radius of the circumcircle (radius 0.5√10), and *DT* is that of the smaller circle (radius *r*), *MD* is the distinction in lengths of the radii, or 0.5√10 – *r*.

Assemble tangent radii to *AB* and *AC* from circle *D*. Every radius is *r*. Since angle *A* is a proper angle, there’s a sq. fashioned between *A* and *D* with facet lengths equal to *r*. Thus we will calculate the horizontal and vertical distances between *D* and *M* as |*r* – 0.5| and |*r* – 1.5|. We then have a proper triangle *MED* with legs *ME* = |*r* – 1.5| and *ED* = |*r* – 0.5| and hypotenuse 0.5√10 – *r*. We thus have:

(*r* – 1.5)^{2} + (*r* – 0.5)^{2} = (0.5√10 – *r*)^{2}*r*^{2} – 3*r* + 2.25 + *r*^{2} – *r* + 0.25 = 2.5 – *r*√10 + *r*^{2}*r*^{2} – 4*r* + *r*√10 = 0*r*(*r* – 4 + √10) = 0

Since *r* > 0 was given, the one resolution is *r* = 4 – √10 &approx 0.84.

**References**

2022 JEE Superior paper 1 reply sheet (see Q.8)

https://jeeadv.ac.in/paperwork/jeeadv-2022-paper1.pdf

Unacademy Atoms video options to paper 1 (Q8 is round 59:00)

https://youtu.be/prgX_Wu_O6A?t=3544