Sigma Percentile
JEE Main 2019 (09 April Shift 1)
LEVELJEE Main

Animated Solution for Mathematics - Circles: If a tangent to the circle intersects the coordinate axes at distinct points and , then the locus of the mid-point of is

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Visualized Solution

  • Given circle:
  • Let the tangent touch the circle at a point .

  • The tangent intersects the coordinate axes at distinct points and .

  • Equation of tangent at to is .
  • Since , the equation is:

  • For (x-intercept), set . So, .
  • For (y-intercept), set . So, .

  • Let the midpoint of be .

  • Using the midpoint formula:

  • Rearranging the equations to isolate parameters:

  • The point of tangency lies on the circle .
  • Therefore, .

  • Substitute and into the constraint:

  • Expanding the squares:
  • Multiply the entire equation by :

  • Take the LCM on the left side:
  • Cross-multiply to get:

  • Replace with to get the locus:
  • Rearranging gives:

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

Imagine you are standing in a coordinate plane. At the center, there sits a perfect, serene unit circle, defined by the equation .
Now, imagine a line—a tangent—that kisses this circle at a single, precise point . As this line rotates around the circle, it carves out intercepts on the and axes, which we label as points and .
Our goal is to track the heartbeat of this system: the midpoint of the segment . As the tangent dances around the circle, we seek the path traced by this midpoint.

Capturing the Tangent

To begin, we must pin down the equation of our moving tangent. For any point on the circle , the equation of the tangent line is given by:
To find where this line hits the axes, we set one coordinate to zero. For the -intercept , we set , yielding . Thus, .
Similarly, for the -intercept , we set , yielding . Thus, .

The Midpoint's Path

Now, we define our midpoint . By the midpoint formula, we have:
These equations bridge the tangent's position and the midpoint's location. Rearranging these, we express the original tangency point in terms of our midpoint coordinates:

The Constraint of the Circle

The point is a prisoner of the circle and must satisfy the constraint . This constraint dictates the entire motion of the midpoint.
By substituting our expressions for and into this constraint, we obtain:

The Final Unveiling

Expanding the squares, we get:
Multiplying the entire equation by , we arrive at:
Taking the common denominator, we have . Cross-multiplying yields .
Finally, replacing with the general coordinates , we reach our destination:
This is the equation of the locus. You have successfully mapped the path, turning a dynamic, moving system into a static, elegant expression of geometry.

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