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2006-04-17 20:43:20

Hey thanks I can see what you have done. I wasn't quite sure what it meant by the 'positive quadrant'.

Thanks for explaining the process!

Regards

yonski
2006-04-17 06:41:19

If you rearrange the equation into

you can see what you're doing a bit more, and just pick the coordinates of the centre and the size of the radius straight off. You were right with the answers though.

I'm not totally sure what you mean in part (b). Surely there could be any number of possible tangents to the circle? If you're asking for the points at which the x-axis intersects the circle, then this can be calculated by solving the simultaneous equation:

So all you need to do is substitute the value of y into the equation for the circle:

Therefore the circle intersects the x-axis at points (√5 + 1,0) and (-√5 + 1,0). I think.

2006-04-15 01:20:08

I have done part a) of the question which need checking & need help on part b)

a) Find the centre and radius of the following circle:

2x^2 + 2y^2 + -4x + 10y = 8

radius = 3.3 centre = (1, -2.5) << this is what i got

b) Find the equation of the tangent to the circle when the circle intersects the x-axis in the positive quadrant.