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C2 co-oridnate geomtery

the line y=0 is a tangent to the circle (x-8)^2+(y-a)^2=16 find the value of a
how would I go about doing this first I thought find where the intersect, so sub y=0 into the equation of the circles but that gave me
x^2-16x+48+a^2???
method plz, don't care about the answer
Reply 1
Avatar for qgujxj39
qgujxj39
17
What is the gradient of the line y = 0?

What does the fact that this is a tangent tell you that dydx\frac{dy}{dx} equals for the circle at the point where y = 0?
you know that at (8,0) is where the tangent is below the centre
You also know the radius is 4
therefore the y co-ordinate is 4 above (8,4)
therefore a = 4
i think
Reply 3
Avatar for sa55afras
sa55afras
First I'm substituting, like you said:

Unparseable latex formula:

[br](x-8)^2+(y-a)^2=16[br]\par (x - 8)^2+(-a)^2=16[br]\par (x - 8)^2+a^2=16[br]\par (x - 8)^2=16-a^2[br]



So, we know there's only one intersection point, because it's a tangent. Therefore, we know (x−8)2=16−a2(x-8)^2=16-a^2 (obtained above) has exactly one solution, regardless of the value of x. This means that 16−a2=016-a^2=0 because, if 16−a2<016-a^2<0 there would be no solutions and if 16−a2>016-a^2>0 there would be two solutions.

Does that make sense? Hope it helped :smile:
Reply 4
Avatar for qgujxj39
qgujxj39
17
sa55afras' way is very good, actually. Raminder's is also good, although it assumes the circle is above y=0 rather than below.
tommm
sa55afras' way is very good, actually. Raminder's is also good, although it assumes the circle is above y=0 rather than below.

that is true but the square root of 16 is plus or minus 4. So how would you get around this problem?
Reply 6
Avatar for qgujxj39
qgujxj39
17
Raminder1992
that is true but the square root of 16 is plus or minus 4. So how would you get around this problem?


If you instead say that the tangent is at the top of the circle, then the same method gets you a = -4.
Reply 7
Avatar for sa55afras
sa55afras
tommm
If you instead say that the tangent is at the top of the circle, then the same method gets you a = -4.


In fact there are two circles for which the condition is true, right?
Reply 8
Avatar for qgujxj39
qgujxj39
17
sa55afras
In fact there are two circles for which the condition is true, right?


Yup.
tommm
If you instead say that the tangent is at the top of the circle, then the same method gets you a = -4.

thanks:biggrin:
Reply 10
Avatar for f45
f45
OP
2
umm the answer is 5 :s

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