Using Midpoint And Distance Formulas: Fill & Download for Free

(Sorry I had uploaded the wrong picture earlier)I saw many of you guys claiming that this was the toughest question in JEE Advanced 2019 and to me that was a surprise. I have always felt coordinate geometry as a mind numbing exercise and I guess it is the same reason that you guys thought it was difficult.I will be attaching multiple pictures just for the better understanding of the readers but in the exam it could have been done much more quickly and in much less space.First of I draw both the circles and the line joining their centres as follows:Now since the centre of the third circle lies on the green line above and both the circles lie within that circle and also it touches both of them, no big deal in knowing the locations of M and N.And thus the system should look something like:(Just a little clarification, I have I committed a mistake of interchanging M and N in some of the the pictures that follow. I don’t wish to make the effort of uploading these again so please bear with me.)Now let’s find the radius and centre of the circle C3. Computing radius was fairly straightforward. Just look at the picture below. Use the radius of C1, C2 and distance between their centres which again was fairly easy to compute as it is a commonly used Pythagorean triplet.So the diameter of C3 is 3 + 5 + 4 = 12 which implies the radius is 6.Now coming to the centre. That is also a simple idea and an easy calculation. Now that you know the radius, you have reached to the following figure:Now observe that the centre of C3 divides the line joining the centres of C1 and C2 in 3:2. So the natural idea should be to use the section formula. And wait one of the points is origin so applying the section formula is much simpler.So the coordinates of the centre are (1.8, 2.4)(From this point onwards, I will be using the correct labeling of M and N till now it didn’t anyway matter.)Next I find the equation of common chord. That again is simple. Just take the difference of the equations of both the circles. So the equation of common chord is[math]6x + 8y = 18[/math]which can be simplified to[math]3x + 4y = 9[/math]Next using geometry or coordinate geometry, I know that this line is perpendicular to the line joining the centres of the circles C1 and C2.Now I find the perpendicular distance between the centre of C1 and the common chord using the following formula:[math]\dfrac{\left| 3 \times 0 + 4 \times 0 [/math][math]-[/math][math] 9 \right|}{\sqrt{3^2+4^2}}[/math][math]= 1.8[/math]Now you use the Pythagoras theorem again as follows:This too was the same Pythagorean triplet just multiplied by a factor of 3/5.And using geometry we know that length of XY will be 4.8 cm. Now we use same idea to find the length of ZW.[math]\dfrac{\left| 3 \times 1.8 + 4 \times 2.4 [/math][math]-[/math][math] 9 \right|}{\sqrt{3^2+4^2}}[/math][math]= 1.2[/math]So after using Pythagoras theorem, we got ZW’s length as:[math]= 2.4 \sqrt{6}[/math]which means ZW’s length will be:[math]= 4.8 \sqrt{6}[/math]Now coming to area of the triangle MZN, see the following figure:The base and height have been highlighted in the yellow color above and use the following formula to compute the area[math]\text{Area} = \dfrac{1}{2} \times \text{base} \times \text{height}[/math]which turns out to be[math]\dfrac{1}{2} \times \dfrac{1}{2} \times \text{ZW} \times 12[/math]Now see the next figure:[math]\text{height} = 1.8 + 3 = 4.8[/math][math]\text{Area} = \dfrac{1}{2} \times \text{ZW} \times \text{height}[/math][math]\text{Area} = \dfrac{1}{2} \times \text{ZW} \times 4.8[/math]which means[math]\dfrac{\text{Area of triangle MZN}}{\text{Area of triangle ZMW}} = \dfrac{\dfrac{1}{2} \times \dfrac{1}{2} \times \text{ZW} \times 12}{\dfrac{1}{2} \times \text{ZW} \times 4.8}[/math]which is[math]\dfrac{5}{4}[/math]Now we need to find the common tangent of C1 and C3 which again is pretty straightforward. It will be perpendicular to the line joining the centres of the three circles because any line that passes through the centre of a circle is a normal. So we have the slope and now we only need the coordinates of point M. Now recall that earlier we had seen that the centre of C3 was at a distance of 3 units from centre of C1 and so is M and all three lie on the same line. This means that centre of C1 is the midpoint of line joining M and centre of C3.Hence, coordinates of M are (-1.8, -2.4).[math]\text{Slope of line joining the centres is } \dfrac{4}{3}[/math][math]\text{And so slope of tangent passing through M will be } \dfrac{-3}{4}[/math]So equation of the tangent will be[math]3x + 4y + 15 = 0[/math](You could also get this by taking the difference of the equations of C1 and C3 since it is the common tangent and then not even this bit of thinking was required.)which can also be written as[math]x = -\dfrac{4}{3}y [/math][math]-[/math][math] 5[/math]And we know that the equation of a tangent of the parabola[math]x^{2} = 4ay \text{ is} [/math][math]x = my + \dfrac{a}{m}[/math]Since the equation of the parabola given is[math]x^{2} = 8 \alpha y[/math][math]\text{We get } a = 2 \alpha[/math]So the equation of the tangent now becomes[math]x = my + \dfrac{2 \alpha}{m}[/math]On comparing both these equations of tangents we get[math]m = -\dfrac{4}{3}[/math][math]\dfrac{2 \alpha}{m} = -5[/math][math]\text{which gives } \alpha = \dfrac{10}{3}[/math]There was hardly any difficult calculation or trick or difficult concept being applied here. This was slightly longish but still both the parts could have been completed well within 5 minutes and the weightage of mathematics section in each paper was 62 marks and this was a 6 mark question so if you allot your time uniformly to questions based on their weightage then this question deserved around 6 minutes of effort so not that big deal really. And if this is the toughest question, then mind you the paper was not difficult at all, slightly longish maybe but not difficult. If anything, the following questions were more difficult than the question mentioned above:The first question was pretty straightforward and not even worth a minute and a half if one would have thought very naively but there was no formal concept that is in the syllabus of JEE Advanced that could have been applied to this. The second question is about fibonacci numbers but again this could have been done as this only involved algebraic manipulations but some of those manipulations were really tricky and would have been hard to come up with in the exam environment. And I have not yet reviewed the entire paper so I am not saying that these are the only tough questions but the one mentioned above is certainly not the toughest.