Find the differential equation of family of circles each of which touch the lines $ y=x $ and $y=-x$

Question

A family of circle has $3$ arbitrary constants if I write it as $x^2+y^2+2fx+2gy+h=0$. So since I have two conditions I can reduce this to an equation with 1 arbitrary constant. Then I will get a first order ODE. But how do I do this? That is change the equation to one with only one arbitrary constant. If I substitute $y=x$ and $y=-x$ then subtracting the two easily yields $g=0$. But it doesn't tell me anything about f and h. What can I do for that? Please let me know if what I'm thinking is correct. Thank you. Thank you.

Answer 1

The circle equation is $$ (x-M_x)^2 +(y-M_y)^2=R^2 $$ with tangent equation at some point $(x_0,y_0)$ on the circle $$ (x_0-M_x)(x-x_0)+(y_0-M_y)(y-y_0)=0 $$ One of the tangents shall be $y=x$, thus also $y_0=x_0$inserting and comparing linear and constant coefficients one finds $$ (x_0-M_x)+(x_0-M_y)=0 $$ which identifies the coordinates of the tangency point as $x_0=y_0=\frac12(M_x+M_y)$ which in turn requires for the circle equation $$ (M_y-M_x)^2=2R^2 $$

From the other tangent $y=-x$ one finds similarly $$ (x_0-M_x)+(x_0+M_y)=0\iff x_0=-y_0=\frac12(M_x-M_y)\\\implies (M_x+M_y)^2=2R^2 $$ For the center points one gets thus the $4$ solutions $M_y=0$, $M_x=\pm\sqrt2 R$ and $M_x=0$, $M_y=\pm\sqrt2 R$.

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Take the first case and compute the $x$ derivative of the circle equation $$ 2(x-\sqrt2 R)+2y(x)y'(x)=0 $$ and insert in the original circle equation eliminating $R$, $$ (yy')^2+y^2=\frac12(x+yy')^2\iff (yy'-x)^2 +2y^2-2x^2=0 $$

Answer 2

To derive the circle equation beginning from your general circle form:

$$ x^2+y^2 + 2 f x + 2 g y + h = 0 \tag {1} $$

Put $x= +y$ and simplify, you get

$$ 2 x^2 + (2 f+2 g)x +h =0 , \quad x^2+ (f+ g) x + h/2 =0 \tag{2}$$

Discriminant should vanish when we have tangency (but not secancy :), which cuts at two points like the situation at H).

$ x^2=y^2$">

$$ (f+g)^2 = 4 * h/2 = 2 h \tag{3} $$

Similarly for $ x= -y $ we have

$$ (f-g)^2 = 4 * h/2 = 2 h \tag{4} $$

Solving (3),(4)

$$ g=0,\, f = \sqrt {2 h} = \sqrt {2} u \tag {5} $$

Plug back into (0) and re-write

$$ x^2 + y^2 - 2 u\sqrt 2 + u^2 =0 \tag{6} $$

$$ ( y - \sqrt 2 \,u)^2 +x^2 = u^2 \tag{7}$$

Similarly for the other set of circles centered on x-axis.

There are these two sets we see with the same tangency loci $ x= \pm y$ also called envelopes, with full symmetry about both axes of $x$ and $y$. So change sign before $\pm$ symbol is admissible.

$$x^2+(y−u √2)^2=u^2,x^2+(y−v √2)^2=v^2 \tag{8} $$

Differentiate and eliminate the constants $(u,v)$ in either case. The latter part I leave it to you, others also indicated it in their answers.

EDIT1:

Radius of circle is u, power $u^2$ $$x^2 +(y-\sqrt 2 u)^2= u^2 $$ Differentiate once and cancel 2 $$ x+y'(y-\sqrt 2 u) =0 $$ Differentiate once again to obtain second order DE $$ 1 +(y-\sqrt 2 u) y^{''}+y'^2 =0 $$

$$ y +\frac{1+y'^2}{y''}= \sqrt2 u $$

To remove RHS third time differentiate and simplify

$$ y' + \frac{y''(2 y'y'') -(1+y'^2) y'''}{y''^2} =0 $$

$$ y'''= \frac{ 3 y'y''^2}{1+y'^2}$$