y = 2x is the chord of the given circle.
It satisfies the equation of a given circle.
Substituting y = 2x in x2 + y2 – 10x = 0, we get
⇒ x2 + (2x)2 – 10x = 0
⇒ x2 + 4x2 – 10x = 0
⇒ 5x2 – 10x = 0
⇒ 5x(x – 2) = 0
⇒ x = 0 or x = 2
When x = 0, y = 2x = 2(0) = 0
∴ A = (0, 0)
When x = 2, y = 2x = 2 (2) = 4
∴ B = (2, 4)
End points of chord AB are A(0, 0) and B(2, 4).
Chord AB is the diameter of the required circle.
The equation of a circle having (x , y ) and (x , y ) as end points of diameter is given by
(x – x1 ) (x – x2) + (y – y1) (y – y2) = 0
Here, x1 = 0, y1 = 0, x2 = 2, y2 = 4
The required equation of the circle is
⇒ (x – 0) (x – 2) + (y – 0) (y – 4 ) = 0
⇒ x2 – 2x + y2 – 4y = 0
⇒ x2 + y2 – 2x – 4y = 0
