Equation of the family of circle which touch the pair of lines $x^2-y^2+2y-1=0$

Question

The question was to find the equation of the family of circles which touches the pair of lines, $x^2-y^2+2y-1 = 0$

So I tried as follows:-

The pair of lines is, by factoring the given equation, $x^2-y^2+2y-1 = 0$ is $x+y-1=0$, and $x-y+1=0$. These are tangent to the required circle, so the center of the circle, (let that be $(h,k)$) must lie upon the angle bisector of these two lines.

Also, the distance of the center $(h,k)$ from these two lines must equal to the radius (assuming it to be $r$).

How do I use these to get the required equation?

Answer 1

Let $(u, v)$ is a center of a circle.

Since both lines $x+y-1 = 0$ and $x-y+1 = 0$ are the tangent line, the distance from $(u, v)$ to these lines must be the same.

By using the formula of the distance from a point to a line, we have the following equation:

$$ r = \frac{|u-v+1|}{\sqrt{2}} = \frac{|u+v-1|}{\sqrt{2}} \Leftrightarrow |u-v+1| = |u+v-1|. $$

Here, $r$ is the radius of a correponding circle.

There are four possible scenarios:

Scenario 1.

$$ \left\{ \begin{aligned} u-v+1 &> 0 \\ u+v-1 &> 0 \\ u-v+1 &= u+v-1 \end{aligned} \right. \Leftrightarrow (u > 0, v = 1) $$

The radius of such a circle equals to $$ r = \frac{|u|}{\sqrt{2}}, $$

and the corresponding family of circles is:

$$ (x-u)^2 + (y-1)^2 = \frac{u^2}{2}, \text{ }u > 0. $$

Scenario 2.

$$ \left\{ \begin{aligned} u-v+1 &< 0 \\ u+v-1 &> 0 \\ -(u-v+1) &= u+v-1 \end{aligned} \right. \Leftrightarrow (u = 0, v > 1) $$

The radius of such a circle equals to $$ r = \frac{|v-1|}{\sqrt{2}}, $$

and the corresponding family of circles is:

$$ x^2 + (y-v)^2 = \frac{(v-1)^2}{2}, \text{ }v > 1. $$

Scenarios 3 and 4 I left to you.

Scenario 3.

$$ \left\{ \begin{aligned} u-v+1 &> 0 \\ u+v-1 &< 0 \\ u-v+1 &= -(u+v-1) \end{aligned} \right. $$

Scenario 4.

$$ \left\{ \begin{aligned} u-v+1 &< 0 \\ u+v-1 &< 0 \\ -(u-v+1) &= -(u+v-1) \end{aligned} \right. $$

Answer 2

I propose this approach (I will find circles in the right sector, for the others sectors the computation is really similar):

Choose a point $A_a=(a,a+1)$ ($a>0$) that lies in the line $r=\{(x,y) \ | \ y=x+1\}$.

Now draw the line passing through the point $A_a$ and it is orthogonal to the line $r$: with a simple computation you find $s_a=\{(x,y) \ | \ y = -x + 2a+1\}$. In this line will lie the radius of the circle we are looking for.

Take the intersection point between $s_a$ and the right bisector (that has equation $y=1$). You find the point $C_a = (2a,1)$. This will be the center of our circle.

Find the distance $\overline{A_aC_a}$ that is $r_a = \sqrt{(a-2a)^2 + (a+1-1)^2}=a\sqrt 2$; this is the length of radius.

Now we have all the element to write the equation of the circle; the points lie in the circle are points $P$ such that $\overline{PC_a} = r_a^2$ so the equation is: \begin{gather} (x-2a)^2+(y-1)^2 = (a\sqrt 2)^2\\ x^2+y^2-4ax-2y+2a^2+1 =0 \end{gather}

Answer 3

As yuou noted, the centre of the circle must lie on $y=1 \vee x=0$.

In the first case, we consider the centre to be $C(x_c,1)$ and we have: $$(x-x_c)^2+(y-1)^2=r^2$$ The radius is the minimal distance from $C$ to the line $y=x+1$ or $y=-x+1$. So we have: $$r=\frac{|x_c|}{\sqrt{2}}$$ Substituing, we have: $$(x-x_c)^2+(y-1)^2=\frac{x_c^2}{2}$$

In the second case, we consider the centre to be $C(0,y_c)$ and we have: $$x^2+(y-y_c)^2=r^2$$ Using the method expained before, we arrive at: $$x^2+(y-y_c)^2=\frac{(y_c-1)^2}{2}$$

Answer 4

EDIT1:

Factorizing gives two straight lines

$$y-1=-x,\, y-1=+x $$

Depressing the pair of straight lines by $1.0$ for all y-coordinate results the pair at $\pm 45^{\circ}$ slope easy to handle:

$$ (x+y)(x-y)=0 $$

whose touching lines can be readily rough sketched as shown below with variable (x, y) axes displacements $(h,k)$ for location of circle centers.

Now elevate the lines back by same amount $1$

$$y\rightarrow y+1 $$

to obtain required new $(h,k)$ parameter set of circle equations touching the line pair:

$$ (x-h)^2 + (y-1)^2 = h^2/2; \quad x^2+(y-1-k)^2= k^2/2.$$

Answer 5

1) $y=-x+1$; 2)$y=x+1$;

Both lines have $y-$intercept $1$, and intersects at $(0,1).$

2) Angle bisectors :

$y=1$; and $x=0$;

Note: The lines make angles of $\pm 45°$ with the horizontal.

3) For the first quardrant and $y=1$:

The centres of the circles are at $x=t \ge 0$, $y=1$ i.e $C(t,1)$.

$r=t \sin 45°=(1/2)√2t$.

$(x -t)^2-(y-1)^2=(1/2)t^2$;

4) Consider centres of the circles $C(0,t+1)$, $t\ge 0$, i.e. along $y-$axis, $y \ge 1$.

$x^2+(y-(t+1))^2 =(1/2)t^2$.

Can you finish?

Answer 6

The point-line distance formula for the point $(h,k)$ and the line $ax+by+c=0$ is

$$d= \frac{|ah+bk+c|}{\sqrt{a^2+b^2}}$$

Apply the distance formula to the circle centers $(0,p)$, $(p,1)$ on the respective angle bisectors and the tangent line $x\pm y\mp1=0$ to calculate the two sets of radii,

$$r= \frac{|\pm p \mp1|}{\sqrt{2}}= \frac{|p-1|}{\sqrt{2}},\>\>\>\>\> r= \frac{|p \pm 1\mp1|}{\sqrt{2}}= \frac{|p|}{\sqrt{2}} $$

Thus, the two families of the circles are

$$x^2+(y-p)^2= \frac12(p-1)^2,\>\>\>\>\>(x-p)^2+(y-1)^2= \frac12p^2$$