Proof question with Tangents and Circle

[dxoo]

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!

 

[Harry_the_cat]

What do you know about CT and TQ? What do you know about CR and RQ?
What is the length of CQ?
That should help.

 

[HallsofIvy]

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 \(\displaystyle x^2+ y^2= 25\), the quadratic equation \(\displaystyle x^2+ (a(x- 7)- 1)^2= x^2+ a^2x^2- 14ax+ 49- 2ax+ 14a+ 1= (a^2+ 1)x^2- 16ax+ 50= 0\) must have a double root. That means that \(\displaystyle 256a^2- 200(a^2+ 1)= 56a^2- 200= 0\) so \(\displaystyle a= \pm\sqrt{25/7}\). The two tangent lines are \(\displaystyle y= \sqrt{25/7}(x- 7)- 1\) and \(\displaystyle y= -\sqrt{25/7}(x- 7)- 1\).