Question

Find the equation of the circle which passes through the origin and belongs to the co-axial of circles whose limiting points are (1,2) and (4,3)

Answer 1

Equation of circles with limiting points (1,2) and (4,3) are

(x–1)² +(y–2)² = 0  ⇒ x² +y² –2x–4y+5 = 0

(x–4)² +(y–3)² = 0  ⇒ x² +y² –8x–6y+25 = 0

System of co-axial of circles equation is

x² +y² –2x–4y+5+ λ(x² +y² –8x–6y+25) = 0______________(1)

equation (1) passes through origin

∴ 5+25λ= 0

∴ λ = –1/5 Substituting in (1) we get

4(x² +y² )–2x–14y = 0

2x² +2y² –x–7y = 0