Alternative form of equation of circle?

Question

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?

Answer 1

The equation of $C_2$ is $(x-h)^2+(y-k)^2=r^2$ and $C_1$, is $(x-m)^2+(y-n)^2=r^2$.

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 $(h,k)$.

Answer 2

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.

Answer 3

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:

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.