Finding Intercepts of a Circle Passing Through Origin

The problem asks us to find the lengths of the intercepts cut off by a circle on the x-axis and y-axis. We are given that the centre of the circle is (3, 4) and that the circle passes through the origin (0, 0).

Step 1: Find the Radius of the Circle

Since the circle passes through the origin (0, 0), the distance between the centre (3, 4) and the origin (0, 0) is the radius of the circle. We can use the distance formula:

\(r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Here, \((x_1, y_1) = (3, 4)\) (the centre) and \((x_2, y_2) = (0, 0)\) (a point on the circle).

\(r = \sqrt{(0 - 3)^2 + (0 - 4)^2}\)

\(r = \sqrt{(-3)^2 + (-4)^2}\)

\(r = \sqrt{9 + 16}\)

\(r = \sqrt{25}\)

\(r = 5\)

So, the radius of the circle is 5 units.

Step 2: Write the Equation of the Circle

The general equation of a circle with centre \((h, k)\) and radius \(r\) is:

\((x - h)^2 + (y - k)^2 = r^2\)

Substitute the centre \((h, k) = (3, 4)\) and the radius \(r = 5\):

\((x - 3)^2 + (y - 4)^2 = 5^2\)

\((x - 3)^2 + (y - 4)^2 = 25\)

This is the equation of the circle.

Step 3: Find the x-intercepts

To find the x-intercepts, we set \(y = 0\) in the equation of the circle and solve for \(x\):

\((x - 3)^2 + (0 - 4)^2 = 25\)

\((x - 3)^2 + (-4)^2 = 25\)

\((x - 3)^2 + 16 = 25\)

\((x - 3)^2 = 25 - 16\)

\((x - 3)^2 = 9\)

Take the square root of both sides:

\(x - 3 = \pm \sqrt{9}\)

\(x - 3 = \pm 3\)

This gives two possible values for \(x\):

\(x = 3 + 3\) or \(x = 3 - 3\)

\(x = 6\) or \(x = 0\)

The circle intersects the x-axis at x = 0 and x = 6. The length of the x-intercept is the distance between these two points, which is \(|6 - 0| = 6\) units.

Step 4: Find the y-intercepts

To find the y-intercepts, we set \(x = 0\) in the equation of the circle and solve for \(y\):

\((0 - 3)^2 + (y - 4)^2 = 25\)

\((-3)^2 + (y - 4)^2 = 25\)

\(9 + (y - 4)^2 = 25\)

\((y - 4)^2 = 25 - 9\)

\((y - 4)^2 = 16\)

Take the square root of both sides:

\(y - 4 = \pm \sqrt{16}\)

\(y - 4 = \pm 4\)

This gives two possible values for \(y\):

\(y = 4 + 4\) or \(y = 4 - 4\)

\(y = 8\) or \(y = 0\)

The circle intersects the y-axis at y = 0 and y = 8. The length of the y-intercept is the distance between these two points, which is \(|8 - 0| = 8\) units.

Conclusion on Intercepts

The intercepts cut off by the circle are 6 units on the x-axis and 8 units on the y-axis.

Revision Table: Circle Properties

Additional Information: Circle Geometry

A circle is a fundamental shape in geometry. Understanding its properties like centre, radius, and equation is crucial. Intercepts are points where the circle crosses the coordinate axes. The distance from the centre to any point on the circle is always equal to the radius. If a circle passes through the origin (0,0), this point can be used along with the centre coordinates to determine the radius using the distance formula.

The calculation of intercepts involves substituting the value 0 for the appropriate variable (y for x-intercept, x for y-intercept) and solving the resulting quadratic equation. The roots of this equation give the coordinates where the circle meets the axis. The length of the intercept is the distance between these points of intersection.

In this specific problem, since the circle passes through the origin (0,0), one of the x-intercepts is at x=0 and one of the y-intercepts is at y=0. This simplified the calculation slightly as we only needed to find the other intersection point on each axis.