Question

Show that two direct common tangents and the transverse common tangent to the two circles S ≡ x² + y² − 6x = 0 and S' ≡ x² + y^{2 +} 2x = 0 form an equilateral triangle.

Answer 1

A(3, 0) and 3 are the centre and radius of S = 0 (see Fig). B(−1, 0) and 1 are the centre and radius of S' = 0. Now AB = 4 = 3 + 1  S ⇒ 0 and S' touch each other externally at (0, 0) and x = 0 (i.e., y-axis) is the common tangent at the origin. Let T be the external centre of similitude. Therefore, T divides the line joining the centres A and B externally in the ratio 3:1. That is,

Let y = m(x + 3) be a tangent from T(−3, 0) to S' = 0. Therefore

Hence, the tangents from T(−3, 0) are x − √3 y + 3 = 0 and x + √3 y + 3 = 0. Therefore, the vertices of the triangle are P(0, 3 ), T(−3, 0) and Q(0, − 3 ), which form an equilateral triangle.