Question

Find the equation of the family of circles which touch the pair of straight lines x^{​​​​​2}​​​​​ - y^{​​​​​​2} + 2y - 1 =0

Answer 1

First lets get both the straight lines

Solve for y: -1 + x^2 + 2 y - y^2 = 0 Multiplying both sides by -1: 1 - x^2 - 2 y + y^2 = 0 y = (2 ± sqrt(4 - 4 (1 - x^2)))/(2) = 1 ± sqrt(x^2): so they are y=1+x and y=1-x

now they both are tangents to (x-h)^2+(y-k)^2 = 1/2

condition for tangency: c=±a*sqrt(1+m^2)

=>1=±a*sqrt(1+1)

=>a=±1/sqrt(2)

so eqn is (x-h)^2 +(y-k)^2 = a^2.