# C2 Circle Geometry, can you explain this last step to me please (why is it 160-25)?

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#1

I'm following this on Yahoo Answers and I understand it all up to the point where it says "So as OT must be perpendicular to PT it follows that PT^2 = 160 - 25 = 135.
Hence PT = sqrt(135) = 11.6 to 3 sf. ".

Why must OT be perpendicular to PT? And Why would PT^2 then be 160 - 25 (which is OP^2 - radius^2)? Can you explain both these to me please? I don't know a) how you know OT is perpendicular to PT, and b) why PT^2 is then 160-25?

Can you explain please, and possibly help me with a diagram too if you could?

Thanks a lot!
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3 years ago
#2
(Original post by blobbybill)
I'm following this on Yahoo Answers and I understand it all up to the point where it says "So as OT must be perpendicular to PT it follows that PT^2 = 160 - 25 = 135.
Hence PT = sqrt(135) = 11.6 to 3 sf. ".

Why must OT be perpendicular to PT? And Why would PT^2 then be 160 - 25 (which is OP^2 - radius^2)? Can you explain both these to me please? I don't know a) how you know OT is perpendicular to PT, and b) why PT^2 is then 160-25?

Can you explain please, and possibly help me with a diagram too if you could?

Thanks a lot!
PT is perpendicular to OT because it is a tangent. It is a fact that every tangent line to a circle is perpendicular to its radius at a given point. Here is the proof. Knowing this, the construction below is a right-angled triangle.

The distance OT is just the radius length, and length OP can be worked out using Pythagoras as you know both of these points. Then just find the third side, PT, by applying Pythagoras again.

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#3
(Original post by RDKGames)
PT is perpendicular to OT because it is a tangent. It is a fact that every tangent line to a circle is perpendicular to its radius at a given point. Here is the proof. Knowing this, the construction below is a right-angled triangle.

The distance OT is just the radius length, and length OP can be worked out using Pythagoras as you know both of these points. Then just find the third side, PT, by applying Pythagoras again.

Thank you, that really helps! When they say the tangent starts at point (x,y), how do you know which way that is, whether that point is at the top or the bottom, and whether the tangent will be drawn horizontally or vertically?
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3 years ago
#4
(Original post by blobbybill)
Thank you, that really helps! When they say the tangent starts at point (x,y), how do you know which way that is, whether that point is at the top or the bottom, and whether the tangent will be drawn horizontally or vertically?
That's why you do a sketch and work with the centre of the circle. (8,17) from (4,5) is clearly a bit to the right and some way up the graph. The sketch doesn't have to be accurate since the distance is deduced merely off the information about points.

You can also have the following situation where all the lengths are the same nonetheless
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