Circle gradient question???

You should know that if gradient A is perpendicular to gradient B, then gradient B works out with that coreelating thing a s a product to -1.

Ah, I messed up on the equation of the tangent at B! -.- I was 0.75 out on the intercept. I rushed through it, which altered my results. Yes, 28.403333... Is correct

Because they meet at a right angle

And that is the rule for perpendicular gradients

How did you guys work out the last part bii) ???
Thanks

How did you guys work out the last part bii) ???
Thanks

Once you know the gradient of the tangent at B, you can figure out at which point the line intercepts the x-axis. From this figure out length BC with a bit of Pythagoras. The rest I think you already know let me know if you'd like me to clarify

Using pythagoras work out AD. Then using your equation line find the point C. Then use pythagoras to find BC.

Using pythagoras work out AD. Then using your equation line find the point C. Then use pythagoras to find BC.

Is the gradient to B -3/4
How do I figure out where it intercepts x axis?

Yeah I know Pythagoras to get the sides



AD is 4 because y coord of point A is 4. Then? My equation line to find C?

See above

Is the gradient to B -3/4
How do I figure out where it intercepts x axis?

Yeah I know Pythagoras to get the sides



AD is 4 because y coord of point A is 4. Then? My equation line to find C?

You find the lenght of AD which is 4 very imply using pythagoras.

Then you use your equation of tangent BC, that you found in the previous part. Sub in y=0 in your equation to get point C. Which should be (64/3,0).

Then, like you did to work out AD, use pythagoras to work out the length of BC!! This should give you 40/3.

Length = AD+11.07+BC =4+11.07 +(40/3)=28.4 (d.p)


QED boom.!

.

is it given to us that D is directly below A? also, did you get 15.32 (2dp) for BC?

Don't use selective treatment to quote me. You know full well how I calculated it.

What I meant was you don't need to do a^2+b^2=c^2 in that format. You can just use the equation:

(x1-x2)^2+(y1-y2)^2 (this is all square rooted). And that I said is derived from pythagoras!

Whilst I realise that it is not pertinent to the thread, I think it is important for you to note that you have misunderstood the nature of Pythagoras's Theorem. You have used the theorem to calculate the length, not some derivation of it. The theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides, which is what you have calculated.