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In a problem set I was solving, one of the solutions used the equation of a circle in the form

$$(x-h)^2 + (y-k)^2 + \lambda(ax + by +c) = 0$$

where,

$(h,k)$ is any point on the circle

$ax+by+c = 0 \ $ is the equation of tangent at the point $(h,k)$, and $\lambda$ is a constant evaluated by fitting the equation to another known point on the circle.

I just can't see how we got this equation. Could you please help?

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3 Answers 3

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Without loss of generality, consider the points $P=(h,k)$, $P'=(-h,-k)$ and $Z=(x,y)$ on a circle, whose center is the origin $O=(0,0)$. See the following figure:

enter image description here

It's a well known fact that the three points form a right-angled triangle.

Let's recall that the square of the length of a cathetus ($||Z-P||$) equals the product of the lengths of its orthographic projection on the hypotenuse ($m$) times the length of this ($2||O-P||$).

But $m$ can be calculated by:

$$m= (Z-P)\cdot \frac{(O-P)}{|| O-P ||}$$

Therefore we get: $$ ||Z-P||^2 = (Z-P)\cdot \frac{(O-P)}{|| O-P ||} (2 || O-P ||) \Rightarrow $$ $$(x-h)^2+(y-k)^2=-2(x-h)h-2(y-k)k \Rightarrow$$ $$(x-h)^2+(y-k)^2 + 2h(x-h)+ 2k(y-k) = 0 \quad(1)$$

The relation $(1)$ is therefore a circle equation.

Note that the expression $$2h(x-h)+ 2k(y-k) = 0$$ is an equation of the tangent line at point $P$.

So if you choose a suitable $\lambda$ such that $$2h(x-h)+ 2k(y-k) = \lambda (ax+by+c)$$ you will get: $$(x-h)^2+(y-k)^2 + \lambda (ax+by+c) = 0. \quad(2)$$ And we're done.

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  • $\begingroup$ But in this case always $\lambda$ is equal to $2$, is a particular case? or not? $\endgroup$ Commented May 18, 2014 at 18:45
  • $\begingroup$ @AsdrubalBeltran It's not necessary that $a=h$ and $b=k$.The coefficients $a$ and $b$ can be $a=nh$ and $b=nk$ where $n$ is a non zero real number. $\endgroup$ Commented May 18, 2014 at 18:58
  • $\begingroup$ ha ok I also got the same result but with a different process. $\endgroup$ Commented May 18, 2014 at 19:11
  • $\begingroup$ @AsdrubalBeltran I think you could show your process, since is a different one. It can more comprehensible to OP. $\endgroup$ Commented May 18, 2014 at 19:22
  • $\begingroup$ ok I post, in one moment, thank you. $\endgroup$ Commented May 18, 2014 at 19:37
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The equation of $C_2$ is $(x-h)^2+(y-k)^2=r^2$ and $C_1$, is $(x-m)^2+(y-n)^2=r^2$.

enter image description here

then now, make the transla the center of $C_2$ to center of $C_1$ then: $$C_1=(x-h+m-h)^2+(y-k+n-k)^2=r^2$$ $$C_1=(x-h)^2+2(x-h)(m-h)+(m-h)^2+(y-k)^2+2(y-k)(n-k)+(n-k)^2=r^2$$ $$C_1=(x-h)^2+(y-k)^2+2(x-h)(m-h)+2(y-k)(n-k)=0$$ Note that $$2(x-h)(m-h)+2(y-k)(n-k)=2x(m-h)-2h(m-h)+2y(n-k)-2k(n-k)=0$$ is a equation of line tangent in $(h,k)$.

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  • $\begingroup$ This method has a bit of intuition. I like it. Thanks! You flipped a $h$ vs $k$ in your first equation, by the way $\endgroup$
    – Cheeku
    Commented May 19, 2014 at 0:39
  • $\begingroup$ Yes is a error, now i make the changes, $k$ for $h$ in the first equation. $\endgroup$ Commented May 19, 2014 at 0:46
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Given a point $P=(h,k)$ and a line through it $L=ax+by+c=0$, one can have a family of circles s.t. the line is a tangent at $P$ for the circle. This family is given by $C_k=(x-h)^2+(y-k)^2+kL=0$. You can note that for any $k$ this is a circle which passes through $P$ and has exactly one point $P$ in common with the line $L=0$. (Try solving $C_k=0, L=0$).

If you fix that the circle must pass through another point $Q$ also, then the circle gets uniquely determined. As this is exactly what you're doing when you set $k=\lambda$, you get the equation of that unique circle.

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  • $\begingroup$ Yeah, but my question asks for a proof, preferably geometric, showing why $C_k$ represents a circle. My question is not that I don't understand what it is. I just don't get how we got there. $\endgroup$
    – Cheeku
    Commented May 18, 2014 at 5:04
  • $\begingroup$ Ah. You should then specify that in the question, algebraically it's simpler than all the above. It's not hard to see what quadratic forms represent a circle, you should be able to complete the square and show the form is a circle in this case. $\endgroup$
    – Macavity
    Commented May 18, 2014 at 5:07
  • $\begingroup$ That is a reverse proof. Going back to known relations given this relation is not acceptable. I specifically want to derive the said relation geometrically or from the general form. $\endgroup$
    – Cheeku
    Commented May 18, 2014 at 8:03

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