widdonod1t

2021-12-27

Differential Equations (Families of Curves): Circles with fixed radius r and tangent to the y-axis.

poleglit3

The general equation of a circle with centre at (h, k) and radius r is given by
${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}$
Also, it is given that y axis is tangent to this circle.
This means the perpendicular distance from the centre of the circle to the line $x=0$ is equal to the radius of the circle.
This means $h=r$
Hence, the equation of such a circle becomes
${\left(x-r\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}$
Here, k is hte only variable left.
Hence, to obtain the differential equation, we need to eliminate the variable k
We will do this with the help of differentiation.
Differentiating the curve with respect to x, we get
$2\left(x-r\right)+2\left(y-k\right)\frac{dy}{dx}=0$
$\left(y-k\right)\frac{dy}{dx}=r-x$
$\left(y-k\right)=\frac{r-x}{\frac{dy}{dx}}$
Putting the value of $y-k$ back in the equation we get
${\left(x-r\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}$
${\left(x-r\right)}^{2}+\left(\frac{r-x}{\frac{dy}{dx}}\right)={r}^{2}$
${\left(\frac{dy}{dx}\right)}^{2}{\left(x-r\right)}^{2}+{\left(x-r\right)}^{2}={\left(\frac{dy}{dx}\right)}^{2}\left({r}^{2}\right)$
${\left(\frac{dy}{dx}\right)}^{2}\left({x}^{2}+{r}^{2}-2xr-{r}^{2}\right)+{\left(x-r\right)}^{2}=0$
${\left(\frac{dy}{dx}\right)}^{2}\left({x}^{2}-2xr\right)+{\left(x-r\right)}^{2}=0$
${\left(\frac{dy}{dx}\right)}^{2}=\frac{{\left(x-r\right)}^{2}}{2xr-{x}^{2}}$

Marcus Herman

$⇒{\left(x-a\right)}^{2}+{y}^{2}={r}^{2}$
$⇒2\left(x-a\right)+2y{y}_{1}=0$
$⇒\left(x-a\right)=-y{y}_{1}$
$⇒{y}^{2}{y}_{1}^{2}+{y}^{2}={r}^{2}$

Vasquez

Center is (0,k) and radius is r.
Equation is
${x}^{2}+\left(y-k{\right)}^{2}={r}^{2}$ Equation 1
By differentiating we get,
$2x+2\left(y-k\right)\frac{dy}{dx}=0$
$k=+x\frac{dx}{dy}+y$ Equation 2
Putting the value of K from Equation 1 and 2, we get
$\begin{array}{}{x}^{2}+\left(y-x\frac{dx}{dy}-y{\right)}^{2}={r}^{2}\\ {x}^{2}+{x}^{2}\left(\frac{dx}{dy}{\right)}^{2}={r}^{2}\\ 1+\left(\frac{dx}{dy}{\right)}^{2}=\frac{{r}^{2}}{{x}^{2}}\\ \left(\frac{dy}{dx}{\right)}^{2}=\frac{{x}^{2}}{{r}^{2}-{x}^{2}}\end{array}$

star233

$x=a±\sqrt{{r}^{2}-{y}^{2}}$
Explanation:
To solve the differential equations for circles with a fixed radius r and tangent to the y-axis, we can start by considering the equation of a circle with center (a, b) and radius r, which is given by:
$\left(x-a{\right)}^{2}+\left(y-b{\right)}^{2}={r}^{2}$
Since the circle is tangent to the y-axis, it means that the center lies on the x-axis, which implies b = 0. Substituting this into the equation above, we get:
$\left(x-a{\right)}^{2}+{y}^{2}={r}^{2}$
To find the family of curves that satisfy this equation, we need to solve for x in terms of y. Let's proceed with the solution:
$\left(x-a{\right)}^{2}+{y}^{2}={r}^{2}$
Expanding the square term, we have:
${x}^{2}-2ax+{a}^{2}+{y}^{2}={r}^{2}$
Rearranging the terms, we get:
${x}^{2}-2ax+\left({a}^{2}+{y}^{2}-{r}^{2}\right)=0$
This is a quadratic equation in x, and we can solve it using the quadratic formula:
$x=\frac{-\left(-2a\right)±\sqrt{\left(-2a{\right)}^{2}-4\left({a}^{2}+{y}^{2}-{r}^{2}\right)}}{2}$
Simplifying further, we have:
$x=\frac{2a±\sqrt{4{a}^{2}-4{a}^{2}-4{y}^{2}+4{r}^{2}}}{2}$
$x=a±\sqrt{{r}^{2}-{y}^{2}}$
Therefore, the equation of the family of circles with radius r and tangent to the y-axis is:
$x=a±\sqrt{{r}^{2}-{y}^{2}}$

alenahelenash

Step 1:
To solve the differential equation for circles with a fixed radius, r, and tangent to the y-axis, we can start by considering the equation of a circle with center (a, r) and radius r. The equation of such a circle can be written as:
$\left(x-a{\right)}^{2}+\left(y-r{\right)}^{2}={r}^{2}$
Since the circle is tangent to the y-axis, the x-coordinate of its center, a, will be equal to the radius r. Therefore, we have:
$\left(x-r{\right)}^{2}+\left(y-r{\right)}^{2}={r}^{2}$
To find the family of curves satisfying this equation, we need to eliminate the square root. We can do this by squaring both sides of the equation:
$\left(x-r{\right)}^{2}+\left(y-r{\right)}^{2}={r}^{2}$
Expanding the equation, we get:
${x}^{2}-2xr+{r}^{2}+{y}^{2}-2yr+{r}^{2}={r}^{2}$
Simplifying the equation, we have:
${x}^{2}-2xr+{y}^{2}-2yr=0$
Step 2:
Now, we can factor out r from the terms involving x and y:
$r\left(x-2\right)+y\left(x-2\right)=0$
Factoring out the common factor of (x - 2), we obtain:
$\left(x-2\right)\left(r+y\right)=0$
Setting each factor equal to zero, we have two possible cases:
Case 1: x - 2 = 0
In this case, we find that x = 2. Substituting this value back into the original equation, we get:
$\left(2-r{\right)}^{2}+\left(y-r{\right)}^{2}={r}^{2}$
Simplifying the equation further, we obtain:
$4-4r+{r}^{2}+{y}^{2}-2yr+{r}^{2}={r}^{2}$
${y}^{2}-2yr+{r}^{2}=0$
This equation represents a single point, (r, r), which is the point where the circle is tangent to the y-axis.
Case 2: r + y = 0
In this case, we find that y = -r. Substituting this value back into the original equation, we get:
$\left(x-r{\right)}^{2}+\left(-r-r{\right)}^{2}={r}^{2}$
Simplifying the equation further, we obtain:
$\left(x-r{\right)}^{2}+4{r}^{2}={r}^{2}$
${x}^{2}-2xr+{r}^{2}+4{r}^{2}={r}^{2}$
${x}^{2}-2xr+4{r}^{2}=0$
This equation represents a parabola with its vertex at (r, 0) and axis parallel to the x-axis.
In conclusion, the family of curves representing circles with a fixed radius r and tangent to the y-axis consists of a single point (r, r) and a parabola with its vertex at (r, 0) and axis parallel to the x-axis.

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