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# Maths C1 circle geometry help watch

1. A tangent is drawn from (8,2) to touch the circle x^2+y^2-4x-8y-5=0

A) Find the radius of the circle

This is fairly straight forward, as I know that when I complete the square I will get the radius squared as 25, then square rooting it will give me 5. So radius = 5

By completing the square I get

So the equation of the circle is (x-2)^2+(y-4)^2 = 25
(x-2)+(y-4) = 5

So centre is (2,4) and radius is 5

Part B) Find the length of the tangent

I know that you have to use pythagoras for it and use the points however, I do not fully know understand how to do it. So I was wondering If you could fully explain me how to find the length of the tangent
2. Find out the lenght from the point to the center of the circle. (x1,y1) (x2,y2) has distance ((x1-x2)^2+(y1-y2)^2)^0.5
Now draw a picture and notice where the right angle is
3. Sketch a diagram and think about where you have a right angle
4. Yes you do Pythagoras... the distance between (2,4) & (8,2) is the hypotenuse and the radius is one of the short sides.
5. (Original post by abdul95)
A tangent is drawn from (8,2) to touch the circle x^2+y^2-4x-8y-5=0

A) Find the radius of the circle

This is fairly straight forward, as I know that when I complete the square I will get the radius squared as 25, then square rooting it will give me 5. So radius = 5

By completing the square I get

So the equation of the circle is (x-2)^2+(y-4)^2 = 25
(x-2)+(y-4) = 5

So centre is (2,4) and radius is 5

Part B) Find the length of the tangent

I know that you have to use pythagoras for it and use the points however, I do not fully know understand how to do it. So I was wondering If you could fully explain me how to find the length of the tangent
That is incorrect btw (see bold).

Anyway, you have to plot the points and there will be a triangle on which you can use pythagoras.
6. This is C2?

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