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Give the equation of a circle with the center $ (a,0) $ which is tangent to the line $ y = x $

I now have $ (x-a)^2 + y^2 = r^2 $ but I don't know how to continue.. please help!

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I'm not sure what you're asking here. –  Tarnation Sep 14 '12 at 6:31
1  
My guess is OP wants the circle to be tangent to the line $y=x$. –  Gerry Myerson Sep 14 '12 at 6:34
    
Yes, English is not my native language so I might have translated it wrong.. –  JohnPhteven Sep 14 '12 at 6:35

3 Answers 3

up vote 0 down vote accepted

Let's calculate the intersection of $y=x$ and the circle.

So, $(x-a)^2+x^2=r^2$ as $y=x$,

$\implies 2x^2-2ax+a^2-r^2=0$

As $y=x$ is a tangent of the circle, the roots of the above equation must be same, so that the two points of intersection coincide.

So, $(-2a)^2=4\cdot 2\cdot (a^2-r^2)$ (as the discriminant$(B^2-4AC)$ of $Ax^2+Bx+C=0$ must be $0$ for the roots to be equal)

$\implies r^2=\frac{a^2}{2}$

So, the equation of the circle :$$(x-a)^2+y^2=\frac{a^2}{2}$$

Observe that the value of $x$ is $\frac{2a}{2\cdot 2}=\frac a 2$ (as the equal roots of $Ax^2+Bx+C=0$ are $-\frac{B}{2A}$), which we did not need here.

The relationship between $r,a$ can attained at least in following 3 ways:

Differentiating, $(x-a)^2+y^2=r^2$ wrt $x$, $\frac{dy}{dx}=\frac{a-x}{y}$

(1)Observe that at the point of intersection the gradient is $1$.

So, $\frac{a-x}{y}=1\implies x+y=a$ , but $y=x$, So, $x=y=\frac{a}{2}$

$\implies r^2=(\frac{a}{2}-a)^2+(\frac{a}{2})^2=\frac{a^2}{2}$

(2) $$\left(\frac{x-a}{r}\right)^2+\left(\frac{y}{r}\right)^2=1$$

So, the parametric equation of the circle $(a+r\cos t,r\sin t) $.

$a+r\cos t=r\sin t\implies a=r(\sin t-\cos t)\implies a^2=r^2(1-\sin2t)$

$\left(\frac{dy}{dx}\right)_t=-\cot t$

So, $-\cot t=1$ as at the point of intersection the gradient is $1$,

$\implies \tan t=-1\implies$ $$ \sin2t=\frac{2\tan t}{1+\tan^2t}=-1$$

$\implies a^2=r^2(1-(-1))=2r^2$

(3)Let the point of intersection be $(b,b)$.

The gradient of $y=x$ is 1.

The gradient of the line joining $(a,o), (b,b)$ is $\frac{b-0}{b-a}=\frac{b}{b-a}$

These two lines are perpendicular, so , $1\cdot \frac{b}{b-a}=-1\implies b=a-b\implies b=\frac a2$

So, $r^2=(\frac{a}{2}-a)^2+(\frac{a}{2})^2=\frac{a^2}{2}$

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I think this answer overcomplicates things a little... you do not need calculus to solve this question. –  Ben Millwood Sep 17 '12 at 17:32
    
@BenMillwood, I've used three independent methods to find the relationship between $r$ and $a$, of which (1) does not use calculus. –  lab bhattacharjee Sep 17 '12 at 17:34
    
Ah, right. I still think you are missing the most elementary method (alluded to by Gerry's answer), but I didn't read carefully enough to realise your three approaches were independent. Perhaps you should indicate so in the question. –  Ben Millwood Sep 17 '12 at 17:36

If the circle is tangent to that line, the point of tangency will be $(a/2,a/2)$. Can you see why that is? Can you use that to find $r$?

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Since the line $y=x$ is tangent to the circle implies the $\perp_r $ distance between the center of the circle and the line $y=x$ must be equal to the radius of the circle. (why?)

Now, $r=\perp_r$ distance of $(a,0)$ from line $y=x$ is $|\frac{a}{\sqrt{1^2+1^2}}|=\frac{|a|}{\sqrt 2}$

Thus, equation of circle is $(x-a)^2+y^2=\frac{a^2}{2}$

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