Math Is Fun Forum

Dharshi

Member

Registered: 2006-10-31

Posts: 56

Tangent to circle

For what values of k would the line y = k be tangent to the circle x^2 + y^2 = 36?

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 48,425

Re: Tangent to circle

x^2 + y^2 = 36
Differentiating with respect to to x,
2x + 2ydy = 0.
2x = -2ydy
I dunoo to go any fruther.

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noobard

Member

Registered: 2009-06-07

Posts: 28

Re: Tangent to circle

this can be easily solved by co ordinate geometry

the equation is a circle with center at origin................radius 6
and y=k is a line parallel to the x axis....
making the graph we can clearly see that it will be tangent only at (0,6) and (0,-6).........so k can be +6,-6
...if u dun like this way .....the perpendicular distance from the center of the circle to the tangent is equal to radius....so if y=k is a tangent then (modulus)k =6
which gives the same thing

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Everything that has a begining has an EnD!!!

Magister_Asmodeus

Member

Registered: 2009-05-24

Posts: 5

Re: Tangent to circle

hi! noobard set the easy way. Here is a way that you can use in most exercises of this type.
In order for the line to be tangent to the circle, the distance between the centre of the circle and the line must be equal to the radius. The formula that gives the distance between a point M(x0,y0) and a line Ax+By+C=0 is d=|Ax0+By0+C|/√ (A²+B²)
So by solving the equation d=R (where R:radius) you can find the values of k you want.

riad

Guest

Re: Tangent to circle

the line is tangent to the circle hf it intersects the cirle once, therfore substituting y=k in the circle equation we have x^2+k^2=36, so x^2+k^2-36=0 and this equation has one solution that is to say , for the equation we must have b^2-4(a)(c)=0, i.e 0-4(1)(k^2-36)=0
so k^2-36=0 and k^2=36   ,  k=6 or k=-6
w.b.w
Mr riad zaidan
Al-quds Open University
jenin
Palistine