Question

Prove that the two circles, which pass through (0,a) and (0,–a) and touch the line y = mx+c, will cut or thogonally if c² = a² (2+m²)

Answer 1

Let the equation of the circles be

x² +y² +2gx+2fy+d = 0

This circle passes through the points (0,a) and (0,–a)

∴a² +2fa+d= 0 __________(1) and a² –2fa+d = 0 _________(2)

4fa = 0

∴f= 0 and d = –a²

∴The equation of circle is x² +y² +2gx–a² = 0

Centre of this circle is (–g,0) and radies \(\sqrt{g^2+a^2}\)

Since line y = mx+c touches the circle

∴ \(\left|\frac{-mg+c}{\sqrt{m^2+1}}\right|\) \(\sqrt{g^2+a^2}\)

c–mg = \(\sqrt{g^2+a^2}\) \(\sqrt{m^2+1}\)

Squaring

c² +m² g² –2mcg = g² m² +g² +a² m² +a²

g² +2mcg+a² (1+m²)–c² = 0

It is a quadratic in g

∴ product of the roots g₁ g₂ = a² (1+m²)–c²

Sum of roots g₁ +g₂ = –2mc

Now the equations of the two circles represented are x² +y² +2g₁ x–a² =0 and x² +y² +2g₂ x–a² = 0

These two circles will be orthogonal if

2g₁ g₂ = – a² –a²

g₁ g₂ = –a²

But g₁ g₂ = –c² +a² (1+m²)

∴–c² +a² (1+m²) = – a²

or c² = a² (2+m²)

Which is the required condition