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

Given that we need to 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.

Since the circle passes through origin O. So, the points on x and y - axis which are intersected by the circle are A(4, 0) and B(0, 6).

The mid - point of O(0,0) and A(4,0) is C(2,0) and that of O(0,0) and B(0,6) is D(0,3).

Let us assume that P is the centre of the circle.

From the figure, we can see that the line PC is perpendicular bisector of the chord OA and line PD is perpendicular bisector of the chord OB.

We get the centre of the circle to be (2,3).

We have a circle with centre (2,3) and passing through the point (0,0).

We know that the radius of the circle is the distance between the centre and any point on the radius. So, we find the radius of the circle.

We know that the distance between the two points (x₁, y₁) and (x₂, y₂) is \(\sqrt{(\text x_1-\text x_2)^2+(y_1-y_2)^2}.\)

Let us assume r is the radius of the circle.

We know that the equation of the circle with centre (p, q) and having radius ‘r’ is given by:

⇒ (x - p)² + (y - q)² = r²

Now we substitute the corresponding values in the equation:

⇒ x² - 4x + 4 + y² - 6y + 9 = 13

⇒ x² + y² - 4x - 6y = 0.

∴ The equation of the circle is x² + y² - 4x - 6y = 0.