circle stuff

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[raven axle]

@proper lynx In the Points where x^2-2x+y^2+4y=20 and x+7y=12 Intersect both equations have to hold. That gives you a system of equations that you just have to solve. To do that solve x+7y=12 for x and substitute that into x^2-2x+y^2+4y=20.

[proper lynx]

so simultaneous equations to find x?

[raven axle]

Yes

[proper lynx]

wouldnt that only cancel one of the x terms though

[raven axle]

$x+7y=12$ so $x=12-7y$. Sub that into the other equation to get $(12-7y)^2-2(12-7y)+y^2+4y=20$. As you can see both x terms are gone.

[bleak root]

Schlaumau

[raven axle]

@proper lynx

[proper lynx]

ohhhh

thankyou

[proper lynx]

sorry i have another question, if all the terms are now in y how am i supposed to find 2 points when co-ordinates follow the (x,y) format

[raven axle]

As this is a quadratic equation you will get two values for y. For each of the values you will get you can calculate the corresponding x value with x=12-7y.

[proper lynx]

perfect thankyou

[raven axle]

Maybe start by solving this equation and you will see what I mean.

[proper lynx]

50y^2-178y+100=0

then factorise

?

2(25y^2-89y+50)

i think i did something wrong i only got 1 y value @raven axle

fixed it

quad formula

[raven axle]

$$49y^2-168y+144-24+14y+y^2+4y=50y^2-150y+120=20$$
$$y^2-3y+2=0$$

[bleak root]

Schlaumau

[raven axle]

if you got y=1 and y=2 you are correct.

[proper lynx]

yup

[raven axle]

nice

now just calculate the corresponding x values

[proper lynx]

yeah so just put the y values into the original equation to find the x values?

[raven axle]

yes

But into the second one.

The first one would be unnecesarrily complicated

[candid bison]

what level of maths is this

[raven axle]

Not to high. Just some algebra and in b a bit of calculus.

[proper lynx]

its the first year of the british Alevel maths course

[candid bison]

ah

ye looks like the same as higher maths (scottish firstyear A level equivalent)

[proper lynx]

yeah its the edexcel course that im studying

[candid bison]

nice

Im doing 2 year equivalent

the calculus is a huge jump

[proper lynx]

i loved it ngl

its the only i dont struggle with

[candid bison]

heres an example lol

solution

i use tabular method tho so less working required

[proper lynx]

whats that like 4marks?

[proper lynx]

point A (where y= 1): (5,1)
point B (where y= 2): (-2,2)

[candid bison]

i think 5

[proper lynx]

i have no idea what B is asking for

oops wrong person

[raven axle]

Do you mean here?

[proper lynx]

yah

[raven axle]

Do you know what a tangent is?

[proper lynx]

yeah

a line that touches the outside of a circle

[raven axle]

exactly

b) wants you to find the equations of those two tangents.

[proper lynx]

so y=mx+c

[raven axle]

yes

[proper lynx]

but if the tangent is touching the point the gradient will be 0 right?

[raven axle]

not necesarrily

The gradient will be same as the gradient of the circle in the point where it is touching the circle

how could you find the value of that gradient?

[proper lynx]

y1-x2 /y2-x1

or rise over run

[raven axle]

but you only have one point on the tangent.

Have you learned implicit differentiation?

[proper lynx]

yes

[raven axle]

Then try to use the equation of your circle to find the slope of it in the Point (5,1).

Or maybe in general in the point (x,y)

[proper lynx]

wouldnt that just be like completing the square twice?

[raven axle]

I dont think so.

$2x-2+2yy'+4y'=0$

[bleak root]

Schlaumau

[raven axle]

correct?

[proper lynx]

not sure

[raven axle]

I took x^2-x+y^2+4y=20 and differentiated both sides with respect to x.

[proper lynx]

right

i get it

[raven axle]

Now we need to solve that equation for y'.

can you do that?

[proper lynx]

differentiate both sides with respect to y

[raven axle]

Thats not necesarry.

[proper lynx]

collect the like times

2x-2 = 2yy'-4y'

[raven axle]

$$2x-2+2yy'+4y'=0$$
$$y'(2y+4)=2-2x$$
$$y'=\frac{1-x}{y+2}$$

[bleak root]

Schlaumau

[proper lynx]

so we sub that answer into the y=mx+c format right?

where m = y'

[raven axle]

We should put in values for x and y before we do that.

we can start with (5,1). What would be the gradient in this point?

[proper lynx]

-4/3

[raven axle]

yes.

Now we can sub that in for m

[proper lynx]

ohhh right

so for (5,1)

its 1= (-4/3)5+c

then solve for c?

[raven axle]

yes

[proper lynx]

nice

for c its just doing the same thing we did for A and B right?

and then d is just the (x-h)^2 + (x-k)^2= r^2

[raven axle]

For c set the first tangent equal to the second and solve for x

for d use the distance formula

i have to go now.