# Finding the equation of a curve by using the points of intersection of tangent and normal on the X and Y axis and other information.

Tangent and normal to a curve at any point $P$ meets $X$ and $Y$ axis at points $A, B$ and $C, D$ respectively. Also center of circle through $O$ (origin), $C$, $P$ and $B$ lies on $x + y = 0$. The curve passes through $(1,0)$. Find the equation of the curve.

I assumed the center to be $(a,-a)$ and using the equation of the circle $(x-a)^2 + (y+a)^2 = 2a^2$ found the coordinates of $B$ and $C$ as $(2a,0)$ and $(0, -2a)$. Then I assumed that a point $(x, y)$ on the unknown curve and formed a differential equation for tangent of the curve $\frac{Y-y}{X-x} = \frac{dy}{dx}$ thereby putting the value of coordinates of $C(0,-2a)$ in place of $Y$ and $X$. I found the solution to be $x-1 = \frac{y}{2a}$ by utilizing the point $(1,0)$. Similarly I formed a different differential equation for normal and putting the coordinates of $B$ and got the solution as $y^2 + x^2 - 4ax + 4a = 1$. Now I tried comparing the curves to find the value of $a$ but could not. I think there is some error in my approach.

• A diagram would be helpful to understand your question. – Jens Apr 9 '17 at 21:17