amanvermagmat wrote:
Hello
These are some basics of geometry. As we are given that the two circles are identical, their radii have to be same. And if you observe in the diagram, PM and PN are radii of first circle, while QM and QN are radii of second circle - so these 4 PM, PN, QM, QN must be equal. This makes this quadrilateral a rhombus.
Now if you observe, QM is a tangent to the left circle at the point of contact M. And since P is centre of that circle, P must be perpendicular to PM at point M (again basics). So angle PMQ = 90. similarly angle PNQ is also = 90 and other two angles are also 90.
Thus its a rhombus with all angles 90 degrees, hence its a square.
Thank you for the answer Aman, I'm trying to wrap my head around some of these geometry properties and struggling
Let me try to walk through your explanation and see if I can follow:
So, we know that P and Q are the center of the circles. Are we assuming that M and N are on the circumference of the circle based on the diagrams being drawn to scale? If so then I understand makes PM and PN radius of circle 1, and QM and QN radius of circle 2. Since the circles are identical these line segments are all identical in length, meaning we have a rhombus.
Ok, I'm following so far now that you've explained it. But I'm still not able to figure out how we know that the angles are 90. Are you saying any line that is a tangent to the circle will always be perpendicular to the center? If that's true than why do we need our two additional pieces of information at all? Couldn't we establish the shape is a square by knowing the circles are equal size and that M & N are the points where the circles intersect?
Thank you for helping me understand these concepts
Yes, your first para is completely correct. So you are perfectly clear till the point that we have a rhombus here.
As per your second para, yes, you have raised a wonderful point. Since we already know that 4 sides of this quadrilateral are equal (both circles being identical so radii equal) and two angles are already 90 degrees (radius being perpendicular to tangent at point of contact), so we can already establish that this quadrilateral is a square. Infact, we dont need the two statements given
Kudos for this point.