Question 8 - Jee advanced Math 2022 P1 Questions with Solutions

Let ABC be the triangle with \(AB = 1, AC = 3\) and \(\angle{BAC} = \frac{\pi}{2}\). If a circle of radius \(r > 0\) touches the sides AB, AC and also touches internally the circumcircle of the triangle ABC, then the value of \(r\) is _____. 

Sol :

Let \(C_{1}\) and \(r_{1}\) be the centre and the radius of the circumcircle(circle through the three vertices) of triangle ABC.

Let \(C_{2}\) be the centre of the circle which is touching sides AB and AC of triangle ABC and touching the circumcircle at P(say). From the question, letter \(r\) denotes the radius of this circle. 






The first result that we will use here is :

If two circles touch each other internally(or externally) then their centres and the point of contact of circles are aligned, i.e., points \(C_{1}, C_{2}\) and P are collinear. 

\(\implies r = r_{1} - d(C_{1}, C_{2})\) …..(1)

where \(d(C_{1}, C_{2})\) is the distance between \(C_{1}\) and \(C_{2}\).

The second result that we require here is related to the circumcircle.

In any right angled triangle, circumcircle is always centered at the midpoint of the hypotenuse. In other words, the hypotenuse is diameter of the circumcircle.

\(\implies BC\) is diameter

\(\implies \frac{1}{2}BC\) is radius \(r_{1}\)

By Pythagoras’ theorem,

\(BC = \sqrt{1^{2} + 3^{2}}\)

\(\implies r_{1} = \frac{1}{2}BC = \frac{\sqrt{10}}{2}\)

To find \(d(C_{1}, C_{2})\), let’s re-imagine the figure in a cartesian coordinate plane with vertex A at the origin, vertex B at (0, 1) and vertex C at (3, 0). 


\(C_{1}\) is the midpoint of B(0 ,1) and C(3, 0).

\(\implies\) coordinates of \(C_{1}\) are \((\frac{0 + 3}{2}, \frac{1 + 0}{2})\)

i.e., \((\frac{3}{2}, \frac{1}{2})\)

Let M and N be the points where the inner circle touches the sides AB and AC. Connect \(C_{2}\) with M and N. 



In quadrilateral \(C_{2}\)MAN, more than two interior angles are \(90^{\circ}\), and adjacent sides \(C_{2}\)M and \(C_{2}\)N are equal(radii of the same circle).

\(\implies\) quadrilateral \(C_{2}\)MAN is a square.

\(\implies MA = AN = C_{2}N = C_{2}M = r\)

\(\implies\) coordinates of \(C_{2}\) are \((r, r)\). 

\(\implies d(C_{1}, C_{2}) = \sqrt{(r - \frac{3}{2})^{2} + (r - \frac{1}{2})^{2}}\)

From (1),

\(r_{1} - r = \sqrt{(r - \frac{3}{2})^{2} + (r - \frac{1}{2})^{2}}\)

\(\implies (r_{1} - r)^{2} = (r - \frac{3}{2})^{2} + (r - \frac{1}{2})^{2}\) 

\(\implies (\frac{\sqrt{10}}{2} - r)^{2} = r^{2} - 3r + \frac{9}{4} + r^{2} - r + \frac{1}{4}\)

\(\implies \frac{10}{4} - \sqrt{10}r + r^{2} = 2r^{2} - 4r + \frac{10}{4}\)

\(\implies r^{2} + (\sqrt{10} - 4)r  = 0\)

\(\implies r(r + \sqrt{10} - 4) = 0\)

\(\implies r = 0\) or \(r = 4 - \sqrt{10}\)

It is given that \(r > 0\).

\(\implies r = 4 - \sqrt{10}\)

               \(\approx 4 - 3.162 = 1.838\)

\(\implies r \approx 1.84\)

Comments

Post a Comment

Popular posts from this blog

Question 7 - Jee advanced Math 2022 P2 Questions with Solutions

Image
 Consider the hyperbola \(\frac{x^{2}}{100} - \frac{y^{2}}{64} = 1\) with foci at \(S\) and \(S_{1}\),  where \(S\) lies on positive x-axis. Let \(P\) be a point on the hyperbola, in the first quadrant. Let \(\angle{SPS_{1}} = \alpha\), with \(\alpha < \frac{\pi}{2}\). The straight line passing through the point \(S\) and having the same slope as that of the tangent at \(P\) to the hyperbola, intersects the straight line \(S_{1}P\) at \(P_{1}\). Let \(\delta\) be the distance of \(P\) from the straight line \(SP_{1}\), and \(\beta = S_{1}P\). Then the greatest integer less than or equal to \(\frac{\beta \delta}{9}\sin{\frac{\alpha}{2}}\) is ______. Sol :  There are a couple of properties of hyperbola that will help us here. (1) The tangent line at a point bisects the angle between the lines connecting the two foci with the point of tangency.  So, in the figure above, tangent at \(P\) bisects the angle \(\alpha\) between the lines joining the foci(\(S\) an...
Read more

Question 6 - Jee advanced Math 2022 P2 Questions with Solutions

 Let \(\beta\) be a real number. Consider the matrix \(\Biggl(\matrix{\beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2}\Biggl)\). If \(A^{7} - (\beta - 1)A^{6} - \beta A^{5}\) is a singular matrix, then the value of \(9\beta\) is _____. Sol:  \(A^{n}\) is a matrix A multiplied to itself n times. Consider \(A^{7}\)            \( = A^{6} \cdot A\)            \( = (A^{5} \cdot A) \cdot A\)            \( = A^{5} \cdot (A \cdot A)\)….. Associative law of matrix multiplication            \( = A^{5} \cdot A^{2}\) \(A^{7} - (\beta - 1)A^{6} - \beta A^{5}\)  \( = A^{5} \cdot A^{2} - (\beta - 1)[A^{5} \cdot A] - \beta A^{5}\) For any constant \(p\) and square matrices \(A\) and \(B\),  \(p (A \cdot B) = (pA) \cdot B = A \cdot (pB)\)  \(\implies = A^{5} \cdot A^{2} - A^{5} \cdot [(\beta - 1) A] - \beta A^{5}\) Also, for any square mat...
Read more

Question 3 - Jee advanced Math 2022 P1 Questions with Solutions

In a study about a pandemic, data of 900 persons was collected. It was found that 190 persons had symptom of fever, 220 persons had symptom of cough, 220 persons had symptom of breathing problem, 330 persons had symptom of fever or cough or both, 350 persons had symptom of cough or breathing problem or both, 340 persons had symptom of fever or breathing problem or both, 30 persons had all three symptoms(fever, cough and breathing problem). If a person is chosen randomly from these 900 persons, then the probability that the person has at-most one symptom is _____. Sol :  Let C be a letter for representing ‘Cough’; F for ‘Fever’; and B for ‘breathing problem’.  i) 190 = n(F) = n(F only) + n(F and C) + n(F and B) + n(all three) ii) 220 = n(C) = n(C only) + n(F and C) + n(C and B) + n(all three) iii) 220 = n(B) = n(B only) + n(F and B) + n(C and B) + n(all three) n(F or C or both) = n(F) + n(C) - n(F and C) - n(all three) (We are subtracting n(F and C) because it is already in...
Read more