Eqn of circle touching the pair of straight line $x^2-y^2+2y-1=0$

Question

I have already found out the separate eqns of the pair of straight lines they are $x+y-1=0$ and $x-y+1=0$ and also their point of intersection of these two i.e. $(0,1)$.

Answer 1

If you mean the equation of circle tangent to lines $x+y-1=0$ and $x-y+1=0$, note that the lines are symmetry about y axis , so it's center has a coordinate like $C(0, a); [1<a<∞] and [-∞<a<1]$ .C is the intersection of a line perpendicular to one of these lines and y axis. Now take an arbitrary point $A(x_A, y_A)$ on one of these lines ; find the equation of perpendicular and hence a. $R=AC$ is the radius of the circle and the equation of a circle tangent at line at point $A(x_A, y_A)$ is:

$x^2+(y-a)^2=R^2$

$R^2=(y_A-a)^2+ x^2_A$

That shows there can be many circle tangent to these line depending on values of a and R.

The two lines are also symmetry about line $y=1$.That is the center of circles can have coordinate like $O(x_O=b , 1)$. You can follow the same method to find center O for any arbitrary point like $A(x_A, y_A)$ on each line. The equation of circle will be:

$(x-b)^2+(y-1)^2=R^2$

Where R can be found as:

$R^2=AO^2= (x_A-b)^2+(y_A-1)^2$

And b can be found by solving a system of equation resulted from intersection of a perpendicular from A and line $y=1$.