# Families of Curves: family of circles with center on the line y= -x and passing through the origin

3 posts / 0 new
danedison
Families of Curves: family of circles with center on the line y= -x and passing through the origin

Find the diff equation of family of circles with center on the line y= -x and passing through the origin.

THANK YOU :)

Jhun Vert

$(x - h)^2 + (y - k)^2 = r^2$

$(x - h)^2 + (y + h)^2 = 2h^2$

$2(x - h) + 2(y + h)y' = 0$

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

$-h(1 - y') = -(x + yy')$

$h = \dfrac{x + yy'}{1 - y'}$

$\left( x - \dfrac{x + yy'}{1 - y'} \right)^2 + \left( y + \dfrac{x + yy'}{1 - y'} \right)^2 = 2\left( \dfrac{x + yy'}{1 - y'} \right)^2$

$\left[ \dfrac{(x - xy') - (x + yy')}{1 - y'} \right]^2 + \left[ \dfrac{(y - yy') + (x + yy')}{1 - y'} \right]^2 = 2\left( \dfrac{x + yy'}{1 - y'} \right)^2$

$\left[ (x - xy') - (x + yy') \right]^2 + \left[ (y - yy') + (x + yy') \right]^2 = 2\left( x + yy' \right)^2$

$(x + y)^2 (y')^2 + (x + y)^2 = 2(x + yy')^2$

Infinitesimal

$$\begin{eqnarray} (x-h)^2 + (y+h)^2 &=& 2h^2\\ x^2 + y^2 - 2hx - 2hy &=& 0\\ \dfrac{x^2 + y^2}{x+y} &=& 2h\\ \dfrac{(2x + 2yy')(x+y) - (x^2+y^2)(1+y')}{(x+y)^2} &=& 0\\ (2x+2yy')(x+y) - (x^2+y^2)(1+y') &=& 0\\ (2xy+2y^2 -x^2 - y^2)y' + (2x^2 +2xy - x^2 - y^2) &=& 0\\ (-x^2 + 2xy + y^2)y' + (x^2 + 2xy - y^2) &=& 0\\ (x^2 - 2xy - y^2)dy - (x^2 + 2xy - y^2)dx &=& 0 \leftarrow Answer \end{eqnarray}$$

## Add new comment

### Deafult Input

• Allowed HTML tags: <img> <em> <strong> <cite> <code> <ul> <ol> <li> <dl> <dt> <dd> <sub> <sup> <blockquote> <ins> <del> <div>
• Web page addresses and e-mail addresses turn into links automatically.
• Lines and paragraphs break automatically.
• Mathematics inside the configured delimiters is rendered by MathJax. The default math delimiters are $$...$$ and $...$ for displayed mathematics, and $...$ and $...$ for in-line mathematics.

### Plain text

• No HTML tags allowed.
• Lines and paragraphs break automatically.