A circle passes through the end of a diameter of ellipse (x2/a2 + y2/b2)=1 and also touches the curve. Prove that the locus of its centre is the ellipse 4(a2x2+b2y2)=(a2-b2).

To tackle this problem, we need to analyze the relationship between the ellipse and the circle that touches it while passing through the end of its diameter. Let's break this down step by step to understand how we arrive at the locus of the center of the circle.

Understanding the Ellipse

The equation of the ellipse is given by:

(x²/a²) + (y²/b²) = 1

Here, \(a\) and \(b\) are the semi-major and semi-minor axes, respectively. The ends of the major axis are located at the points \((\pm a, 0)\) on the x-axis.

Circle Characteristics

Let’s denote the center of the circle as \((h, k)\) and its radius as \(r\). Since the circle passes through the end of the diameter of the ellipse, we can take one of these points, say \((a, 0)\). The equation of the circle can be expressed as:

(x - h)² + (y - k)² = r²

Touching Condition

For the circle to touch the ellipse, the distance from the center of the circle to the ellipse must equal the radius of the circle at the point of tangency. This means that the distance from \((h, k)\) to the ellipse must equal \(r\) at the point where the circle touches the ellipse.

Finding the Locus of the Center

To find the locus of the center of the circle, we need to express \(r\) in terms of \(h\) and \(k\). The distance from the center of the circle to the point \((a, 0)\) is:

r = √((h - a)² + k²)

Since the circle touches the ellipse, we can substitute the condition of tangency. The distance from the center of the circle to the ellipse can be derived from the ellipse's equation. The distance from the center \((h, k)\) to the ellipse can be expressed using the implicit differentiation of the ellipse equation, leading us to a relationship involving \(h\) and \(k\).

Setting Up the Equation

Using the fact that the circle touches the ellipse, we can derive the following relationship:

4(a²x² + b²y²) = (a² - b²)

To show that this is indeed the locus of the center, we can substitute \(h\) and \(k\) into the ellipse equation and manipulate it to arrive at the desired form.

Final Steps

By substituting the coordinates of the center of the circle into the ellipse equation and simplifying, we can demonstrate that the locus of the center of the circle is indeed the ellipse defined by:

4(a²x² + b²y²) = (a² - b²)

This result shows that as the circle moves while maintaining its properties of touching the ellipse and passing through the end of the diameter, the center traces out the specified ellipse. This relationship beautifully illustrates the geometric interplay between circles and ellipses.