locus of the foot of the perpendicular drawn from a fixed point (10,0) on the x axis to any tangent to the circle x²+y²=16

To determine the locus of the foot of the perpendicular drawn from the fixed point (10, 0) on the x-axis to any tangent to the circle described by the equation \(x^2 + y^2 = 16\), we can follow a systematic approach. Let’s break this down step by step.

Understanding the Circle and Tangents

The equation \(x^2 + y^2 = 16\) represents a circle centered at the origin (0, 0) with a radius of 4 units. Any tangent to this circle will be a straight line that touches the circle at exactly one point.

Equation of a Tangent to the Circle

The general form of the equation of a tangent to a circle at the point \((x_1, y_1)\) on the circle is given by:

\(xx_1 + yy_1 = r^2\)

In our case, since the radius \(r\) is 4, this becomes:

\(xx_1 + yy_1 = 16\)

Identifying the Tangent Point

Let’s denote the point of tangency as \((x_1, y_1)\). Since this point lies on the circle, it satisfies the circle's equation:

\(x_1^2 + y_1^2 = 16\)

Finding the Foot of the Perpendicular

Now, we want to find the foot of the perpendicular from the point (10, 0) to this tangent line. The slopes of the radius and the tangent line will help us here. The slope of the radius from (0, 0) to \((x_1, y_1)\) is given by:

\(m_{\text{radius}} = \frac{y_1 - 0}{x_1 - 0} = \frac{y_1}{x_1}\)

Consequently, the slope of the tangent line, which is perpendicular to the radius, is:

\(m_{\text{tangent}} = -\frac{x_1}{y_1}\)

Equation of the Perpendicular Line

The equation of the line passing through (10, 0) and having slope \(m_{\text{tangent}}\) can be expressed as:

\(y - 0 = -\frac{x_1}{y_1}(x - 10)\)

This simplifies to:

\(y = -\frac{x_1}{y_1}x + \frac{10x_1}{y_1}\)

Finding the Locus

The point where this line intersects the tangent line will give us the foot of the perpendicular. To find the locus of these feet, we need to eliminate the parameters \((x_1, y_1)\) from our equations.

Using the Tangent Equation

Substituting \(y = -\frac{x_1}{y_1}x + \frac{10x_1}{y_1}\) into the tangent equation:

\(x \cdot x_1 + y \cdot y_1 = 16\)

We can express this as:

\(x \cdot x_1 + \left(-\frac{x_1}{y_1}x + \frac{10x_1}{y_1}\right) y_1 = 16\)

Which simplifies to:

\(x x_1 - x_1 x + 10x_1 = 16\)

Thus, we get:

\(10x_1 = 16\)

This implies:

\(x_1 = \frac{16}{10} = 1.6\)

Equation of the Locus

Now, substituting back into the equation of the circle, we can find the corresponding \(y_1\):

\(1.6^2 + y_1^2 = 16\)

Calculating \(1.6^2\) gives \(2.56\), leading to:

\(y_1^2 = 16 - 2.56 = 13.44\)

Taking the square root, we find:

Thus, \(y_1 = \pm \sqrt{13.44}\) approximately equals \(3.67\) or \(-3.67\).

Final Locus Equation

Therefore, the locus of the foot of the perpendiculars from the point (10, 0) to any tangent of the circle is a line parallel to the x-axis at \(y = \pm \sqrt{13.44}\) which can be simplified further based on the specific tangent equations.

This geometrical interpretation provides insights into how tangents behave concerning fixed points outside the circle. The locus forms a horizontal line at specific heights, showing consistent distances from the x-axis, reflecting the symmetry of the circle.