Proof question with Tangents and Circle

dxoo

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Jul 16, 2020
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Problem statement:
Tangents are drawn from the point Q(7,-1) to meet the circle with equation x^2+y^2=25 at points R and T. Prove that QRCT is a square, where C is the centre of the circle.



My Approach:
Well since the radius is simply 5, then RC and CT are both 5 units; therefore making it a square. However, this is not a sufficient answer and I'm having difficulty coming up with alternative ideas. I think a good approach may be to find linear equation of both tangents but I don't think there's sufficient information to do that. Any help is always appreciated!
 
Any line through (7, -1) can be written as y= a(x- 7)- 1 for some number a. In order to be tangent to the circle x2+y2=25\displaystyle x^2+ y^2= 25, the quadratic equation x2+(a(x7)1)2=x2+a2x214ax+492ax+14a+1=(a2+1)x216ax+50=0\displaystyle x^2+ (a(x- 7)- 1)^2= x^2+ a^2x^2- 14ax+ 49- 2ax+ 14a+ 1= (a^2+ 1)x^2- 16ax+ 50= 0 must have a double root. That means that 256a2200(a2+1)=56a2200=0\displaystyle 256a^2- 200(a^2+ 1)= 56a^2- 200= 0 so a=±25/7\displaystyle a= \pm\sqrt{25/7}. The two tangent lines are y=25/7(x7)1\displaystyle y= \sqrt{25/7}(x- 7)- 1 and y=25/7(x7)1\displaystyle y= -\sqrt{25/7}(x- 7)- 1.
 
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