Question

Find the equation of the circle which passes through the origin and cuts off chords of lengths 4 and 6 on the positive side of the X-axis and Y-axis respectively.

Answer 1

Let the circle cut the chord of length 4 on X-axis at point A and the chord of length 6 on the Y-axis at point B. 

∴ the co-ordinates of point A are (4, 0) and co-ordinates of point B are (0, 6). 

Since ∠BOA is a right angle. 

AB represents the diameter of the circle. 

The equation of a circle having (x₁ , y₁) and (x₂, y₂) as endpoints of diameter is given by (x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0 

Here, x₁ = 4, y₁ = 0, x₂ = 0, y₂ = 6 

∴ the required equation of the circle is 

⇒ (x – 4) (x – 0) + (y – 0) (y – 6) = 0

⇒ x² – 4x + y² – 6y = 0 

⇒ x² + y² – 4x – 6y = 0