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February 3rd, 2018, 11:53 PM  #1 
Newbie Joined: Dec 2014 From: UK Posts: 11 Thanks: 0  Tangents to Circle
Hi all Can someone help me to find the coordinates of point P please? I understand that the tangents are the same length (because they're both coming from a single external point). Does this mean an isosceles triangle is involved in solving it? Any pointers on the next step would be appreciated. Thanks Jim Last edited by skipjack; February 4th, 2018 at 03:12 PM. 
February 4th, 2018, 12:06 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,755 Thanks: 900 
It should be pretty clear that i) $B=(0,5)$ ii) $P=(P_x,5)$ so we just need to solve for $P_x$ The tangent line at $(3,4)$ will have slope $\dfrac{3}{4}$ (show this) The slope of this line is also $\dfrac{4(5)}{3P_x}= \dfrac{9}{3P_x}$ $\dfrac{9}{3P_x} = \dfrac{3}{4}$ $P_x = 15$ 
February 4th, 2018, 12:41 AM  #3 
Newbie Joined: Dec 2014 From: UK Posts: 11 Thanks: 0 
Hi Romsek That is a brilliant answer which I understand. The only part that I don't understand is how you work out the gradient of the tangent? What is the simplest way of doing that please? Regards Jim 
February 4th, 2018, 01:08 AM  #4 
Senior Member Joined: Oct 2009 Posts: 229 Thanks: 81  What do you know about the tangent line of a circle with respect to the radius of the circle? Use this.

February 4th, 2018, 03:14 AM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,725 Thanks: 985 Math Focus: Elementary mathematics and beyond 
Or use the distance formula and the fact that the tangents are of equal length. $$(x3)^2+81=x^2\implies x=15$$ 
February 4th, 2018, 07:10 AM  #6 
Newbie Joined: Dec 2014 From: UK Posts: 11 Thanks: 0  
February 4th, 2018, 07:13 AM  #7 
Newbie Joined: Dec 2014 From: UK Posts: 11 Thanks: 0  
February 4th, 2018, 12:48 PM  #8 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,725 Thanks: 985 Math Focus: Elementary mathematics and beyond  
February 4th, 2018, 12:59 PM  #9 
Newbie Joined: Dec 2014 From: UK Posts: 11 Thanks: 0 
Thanks


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