The Student Room Group
Reply 1
Revenged
Just need a bit of help with this proof

1) x^2 + y^2 -10y = 0

Prove that the line with equation 4x - 3y + 40 = 0 is a tangent to the equation of the circle given above.

:confused:

You need to find an expression for the x or y coordinate when the line and circle intersect then prove there is only one point of intersection - either by completing the square or evaluating the discriminant.
Reply 2
Gaz031
You need to find an expression for the x or y coordinate when the line and circle intersect then prove there is only one point of intersection - either by completing the square or evaluating the discriminant.


So, let me check i understand what your saying.

First i have to subsitute x = or y = into the circle equation.
Simplify it.
The expression i'm left with should be in the form ax^2 + bx + c = 0.
Then b^2 - 4ac = 0 proves it's a tangent???

Is that right?

(i've tried to do it but couldn't - probably might have made a mess at the simplifying stage)

Also, the other question i'm stuck on is to prove that two circles with different radii touch externally - how do i do this?

I think I can do it if they had the same radii as i think it'll just be the mid-points of the 2 centres of the circles, but am confused with what to do when the radii are different sizes.
Reply 3
Revenged

Also, the other question i'm stuck on is to prove that two circles with different radii touch externally - how do i do this?

I think I can do it if they had the same radii as i think it'll just be the mid-points of the 2 centres of the circles, but am confused with what to do when the radii are different sizes.


yes you've understood Gaz's suggestions fine

you could show that the sum of their radii equals the difference between their centres

EDIT: should say "distance" really not "difference"
Revenged
1) x^2 + y^2 - 10y = 0

Prove that the line with equation 4x - 3y + 40 = 0 is a tangent to the equation of the circle given above

Line: 4x - 3y + 40 = 0
-> 4x = 3y - 40
-> x = (3y - 40)/4

Sub. eqn of line into eqn of circle:

[(3y - 40)/4]^2 + y^2 - 10y = 0
-> (3y - 40)(3y - 40) + 16y^2 - 160y = 0
-> 9y^2 - 240y + 16y^2 - 160y = 0
-> 25y^2 - 400y = 0
-> 25y^2 = 400y
-> 25y = 400
-> y = 16

When y = 16: x = [3(16) - 40]/4 = 2

* Line is tangent to circle as there is only 1 distinct real root of the above quadratic and hence 1 POI.

-> Line is tangent to circle at: (2, 16)
Reply 5
RichE
yes you've understood Gaz's suggestions fine

you could show that the sum of their radii equals the difference between their centres


Thanks :cool:

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