Let the equation of the circles be x 2 + y 2 + 2gx + 2fy + d = 0 .......(i)

 these circles pass through (0, a) and (0, –a)

∴ a 2 + 2fa + d = 0 ......(ii)

and a 2 – 2fa + d = 0 ......(iii)

solving (ii) and (iii), we get f = 0, d = – a 2

put these value of f and d in (i), we get

x 2 + y 2 + 2gx – a 2 = 0 ......(iv)

 y = mx + c touch these circles ⇒ =

⇒ g 2 + (2cm) g + a 2 (1 + m 2 ) – c 2 = 0 ......(v)

equation (v) is quadratic in 'g'

∴ Let g 1 and g 2 are its two roots

∴ g 1 g 2 = a 2 (1 + m 2 ) – c 2

 the two circles represented by (iv) are orthogonal

∴ 2g 1 g 2 + 0 = – a 2 – a 2 ⇒ g 1 g 2 = –a 2 ⇒ a 2 (1 + m 2 ) – c 2 = – a 2

c 2 = a 2 (2 + m 2 ) Hence proved

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