Hi everyone,

now I know this is easy peasy for you guys - but bear with me!

I'll write out the whole question first.

7. A circle with centre C has equation \(x^{2}+y^{2}-4x+12y+15=0\).

a) Find:
i) the coordinates of C;
ii) the radius of the circle.

b) Explain why the circle lies entirely below the x-axis.

c) The point P with coordinates (5,k) lies outside the circle.
i) Show that \(PC^{2}=k^{2}+12k+45.\)
ii) Hence show that \(k^{2}+12k+20>0\)
iii) FInd the possible values of k.

Phew!

I completely understand parts a and b, but c has confused me a little. I've tried thinking about it (alot!) but I am just unsure as to what approach I need for c. It's only worth a couple of marks, so it should be fairly easy - but I'm not seeing what I need to do!

Could you point me in the right direction?

Thanks,

Sparky

 

Have you tried graphing it?

 

Nope, I'll try that now.

 

Just graphed it - but I'm unsure what I'm supposed to do from here.

Sorry about that.

 

Just graphed it - but I'm unsure what I'm supposed to do from here.

Sorry about that.

After you graph it, you should get a better picture of what it looks like. From there, it's mainly just plug and play. Insert the values (k is the y value of the center coordinate; center is (h,k)) in the formula and show that it gives the correct result. Then, find where the circle intersects the Y axis (find possible values of 'k').

I hope this helps. Good luck!
Regards,
Der Strom

 

Should your equation not read?

\({x^2} + {y^2} - 4x + 12y + 15 = 0\)

That is why have -4x + 12x?

 

Good spot - I'll change it now!

 

Okay, I think I'm understanding this more - but what values are P and C?

They are points - not numbers?

Sorry if I'm being a bit - slow.

 

I see how to do it now!

\((CP^{2}=) (5-2)^{2}+(k+6)^{2}

= k^{2}+12k+45 \)

Is that correct?

 

Yes, it's correct. Sorry for the late reply. Did you manage to solve the rest?

 

No problem!

I got the rest pretty quickly - it was mainly the twisted notation which confused me. It suddenly clicked that CP'2' was the radius squared!