2

I am familiar with the following equations

$$S_1 + \lambda S_2=0$$ for two circles

$$L_1 + \lambda L_2=0$$ for two lines $$L_1 + \lambda S_1=0$$ for circle and line

But I never really understood how they worked. Now this is a sample questions

A circles touches the parabola $y^2=2x$ at P $(\frac 12, 1)$ and cuts parabola at vertex V. If centre of circle is a Q, find the radius of circle

The formula used here was

$$(x-\frac 12)^2 + (y-1)^2 +\lambda (2x-2y+1)=0$$

Now it’s easy to see that equation is basically hinting at a curve passing through $(1/2, 1)$ and tangent to $(2x-2y+1)$, but how exactly was the form mat determined? How can we tell if this will us gives a circle? Why was the distance formula used in the first part of the equation? Basically I want to know the process of writing such equations .

Aditya
  • 6,191
  • What other ways can you think of combining $S$ and $L$ so the resulting expression is a quadratic in $x,y$ and has coeff of $x^2=$ coeff of $y^2$ and coeff of $xy=0$? – cosmo5 Feb 08 '21 at 10:01
  • @cosmo5 Well yes, the equation is clearly a circle, but I want to know is why that specific combination works. – Aditya Feb 08 '21 at 10:14

2 Answers2

2

$C_1:\left(x-\frac 12\right)^2 + (y-1)^2 =0$

is the equation of the circle having centre in $P\left(\frac12,1\right)$ and radius $r=0$ and $C_2:2x-2y+1=0$ is the equation of the line tangent to the parabola $\mathcal{C}:y^2=2x$ at $P$

A linear combination of the two $C_1+\lambda C_2=0$ represents all circles tangent to the parabola $\mathcal{C}$ at $P$ (see image (2) below) $$\left(x-\frac 12\right)^2 + (y-1)^2+\lambda(2x-2y+1)=0\tag{1}$$ If the circle passes through the origin, the vertex of $\mathcal{C}$, then substitute $(0,0)$ in $(1)$ to get $\lambda=-5/4$.

The circle we are looking for has equation $$\left(x-\frac{1}{2}\right)^2+(y-1)^2-\frac{5}{4} (2 x-2 y+1)=0$$ simplify $$2 x^2+2 y^2-7 x+y=0$$ center is $\left(\frac74,-\frac14\right)$ and radius $r=\frac{5\sqrt 2}{4}$

$$...$$ enter image description here enter image description here

Raffaele
  • 26,371
  • What is the significance of $r=0$ – Aditya Feb 09 '21 at 18:48
  • @Aditya it is a degenerate circle just like the tangent line https://en.wikipedia.org/wiki/Degenerate_conic – Raffaele Feb 09 '21 at 18:52
  • So it’s family of curve equation basically represents a circle tangent to the degenerate circle and the line? – Aditya Feb 09 '21 at 18:57
  • @Aditya represents ALL circles tangent to the line in that point like in the image 2 I just added to my answer. They are infinite, one for each value of $k$ – Raffaele Feb 09 '21 at 19:05
  • Ok, so in general, if we want a curve to pass through a particular point, we should use $(x-h)^2 + (y-k)^2 +\lambda(C)$? – Aditya Feb 10 '21 at 02:39
  • @Aditya Also $C$ must pass through $(h,k)$ in order to have a family of curves passing through $(h,k)$. Furthermore your equation is incomplete. You should add "$=0$". – Raffaele Feb 10 '21 at 10:02
0

If $S_1(x,y)=0$ and $S_2(x,y)=0$ are the equations of two curves, the equation

$$S_1(x,y)+\lambda S_2(x,y)=0$$ describes a family of curves that contains $S_1$ (when $\lambda=0$) and $S_2$ (when $\lambda=\infty$).

The type of the intermediate curves depend on the formula of the two curves. In particular, if the extreme curves are conics, so are all curves in the pencil. We can also say that if $S_1,S_2$ have common points (intersections), then any curve of the family passes through these points. (See why ?)

Raffaele
  • 26,371