1. ## circles

O is the centre of the small circle. CA and CB are tangents to this circle. The large circle is drawn through A, B and C. CO is produced to E and intersects the small circle at D.
i. Show that O lies on the large circle

ii. Show that CD = (CB x CA) / (CD + 2DO)

2. Originally Posted by deej813
O is the centre of the small circle. CA and CB are tangents to this circle. The large circle is drawn through A, B and C. CO is produced to E and intersects the small circle at D.
i. Show that O lies on the large circle

ii. Show that CD = (CB x CA) / (CD + 2DO)
I can't do it!.
Either there is something missing or mistyped.
The circumcircle ABC completely encloses the small circle with center at O

3. Originally Posted by aidan
I can't do it!.
Either there is something missing or mistyped.
The circumcircle ABC completely encloses the small circle with center at O
Is it not something like this drawing below.
(D & E might be the wrong way round)
Mind you, even with the drawing I cant answer the OPs question,
I can see that no 1 is true but I have no idea how to prove it.

4. Let Q be the point on CO at which the larger circle intersects CO. We want to show that this O.

Since CQ is a diameter of the larger circle, angle CAQ is a right angle (standard theorem in geometry). So as CA is a tangent of the smaller circle, and is perpendicular to AQ, it follows that AQ must be a radius of the smaller circle and so AQ = AO.

So Q = O.

That's the first bit done.

5. Originally Posted by deej813
O is the centre of the small circle. CA and CB are tangents to this circle. The large circle is drawn through A, B and C. CO is produced to E and intersects the small circle at D.
i. Show that O lies on the large circle

ii. Show that CD = (CB x CA) / (CD + 2DO)
I think the picture should look like the attachment. Follow Matt Westwood's advice for part (i). For (ii), notice that CB = CA and CE = CD + 2DO (assuming that E is the point where the line CDO crosses the small circle again). Then the result CD = (CB x CA) / (CD + 2DO) is equivalent to $CD\times CE = CA^2$. This is a standard result which you can prove for example by using the fact that the triangles ADC and EAC are similar.

6. ok thanks i get i.
but with ii. i dont get how CD = (CB x CA) / (CD + 2DO) is equivalent to CD x CE = CA^2

7. Originally Posted by deej813
ok thanks i get i.
but with ii. i dont get how CD = (CB x CA) / (CD + 2DO) is equivalent to CD x CE = CA^2
Multiply out the fraction. Then the equation CD = (CB x CA) / (CD + 2DO) becomes CD × (CD + 2DO) = CB × CA. But you can see from the picture that CE = CD + DO + OE, where DO and OE are both equal to the radius of the small circle. So CE = CD + 2DO. Also, CB = CA (because the two tangents from a point to a circle are always equal).