Equations of two tangents to a circle - A Level Maths

Here is a diagram for the question.

What have you worked out so far?

I know that the circle equation in the form (x-a)^2+(y-b)^2=r^2 is (x+2)^2+(y+6)^2=17
Therefore the centre is (-2,-6) and the radius is √17

Using √(x1-x2)^2+(y1-y2)^2
I can find the length of the line from (5,0) to the centre which is √85

As the tangents meet the circle at 90 degrees you can use Pythagoras to work out the length of the tangent (from (5,0) to where it meets the circle) which is 2√17

You've pretty much got most of it. What are you stuck on?

You could use the usual tangent - circle intersection where the line gradient is to be found using the quadratic discriminant.

Thanks for your help so far. Would you be able to explain this method further? Thank you

Your circle equation is a quadratic and your line is
y = m(x-5)
Sub that into the circle to get a quadratic in x. This must have a single solution if the line is a tangent, so solve the discriminant=0 and that should be a quadratic in m, giving the two gradients.

If you wanted to keep going with your approach, you could have got the gradients from the tan(x+y) formula with a few extra lines.

Hi, I just used the quadratic discriminant method and I ended up getting a polynomial to the power of 4! Upon solving for m I got 0.22 and -0.58. Any ideas where I am going wrong?

Can you upload your working? You should get a quadratic in x,m

This is my working.

Will have a look, but it usually falls out a bit easier.
Can you do it using the tan(x+y) formula? You've got tan(x) and tan(y).

Hi, I've never come across this formula before so not sure what to do. When you say x and y are you referring to the point where the tangent meets the circle? Thanks

This is my working.

Hi,

Would someone be able to help with the following maths question?

A circle has equation x^2+y^2+4x+12y=-23

The lines l1 and l2 are both tangents to the circle, and they intersect at the point (5,0).

Find the equations of l1 and l2, giving your answers in the form y=mx+c.

Thanks in advance.

This is my working.

I'm not saying this is *the* way to do it, but from (x+2)^2 + (m(x-5) +6)^2 = 17, substitute X = x-5 to get:
(X+7)^2+(mX+6)^2 = 17; if you carry on from here you will end up with b^2-4ac being a quadratic in m.
(My gut feeling is that you shouldn't need the substitution, but for sure the algebra gets messy and I didn't really want to spend the time trying to work it through).