stuck on this circle coordinate geometry question

need some help on Q7 please.

Solve for y = 0 to get a quadratic and then use implicit diff to find the gradients. From then, since you have a gradient and point, you can work out the equation of the line.

Posted from TSR Mobile

hello, just wanted to check I was on the right lines.
When I set y=0 and solved the quadratic equation, I got two value for x which were x=5 or x=1. So I presume the points of intersection with the x-axis are (5,0) and (1,0).
After I differentiate the equation I get 2x+2y-2 so do i then put x=5,y=0 and x=1,y=0 into that equation to get the gradient. If thats the case, I got gradient of 8 and 0. Not sure it its right though to have two different gradient with one being a straight line.

Your implicit differentiation is wrong.

$\dfrac{d}{dx} y = \dfrac{dy}{dx}$

Posted from TSR Mob


yh my answers are wrong

What is the answer?

I got the tangents are y = x-5 and y= -x + 1

I will happily explain if they're right!

need some help on Q7 please.

Yeah then find where x is by plugging in y to get the two co-ordinates of where your tangent starts then find the gradient of them from the middle of the circle.

Not sure my answers are right though!



What is the answer please!!

I got the gradient for the (5,0) one as -1 and the other one as 1... did you get the same? :O

does anyone know how we get to these solutions please

It's pretty much what I have it is just they have everything on one side.
Do you know how to get the circle in the form $(x-a)^2+(y-b)2=r^2$?
If you can do that set y=0 to find where it crosses the x axis

ok yes i know how to do that. What I done was i set y=0, to make the equation a quadratic then factorised to get the solutions of x=5 or x=1. So this means that it crosses the x-axis at (5,0) and (1,0), am I right here? I am just stuck on the next part where you have to differentiate the original equation. Any help please.

No need to differentiate. Work out the centre of the circle first.
After you have done that workout the gradient of the line using the coordinates (5,0) and your centre. Do the same for the other point (1,0) and plug into y-y1=m(x-x1).
Tell me if you get stuck

ok thanks you, I was told to differentiate by the first poster to this question