Question

A conic passes through the point (2,4) and is such that the segment of any of its tangents at any point contained between the coordinate axes is bisected at the point of tangency. Then equation of auxiliary circle of the conic is____

A. X² + Y²= 16
B. X² + Y²= 25
C. X² + Y²= 4√2
D. None of these

Answer 1

The equation of the auxiliary circle of the given conic section is x² + y² = 4√2.

Step-by-step explanation:

The equation of an auxiliary circle for a conic section passing through the point (2,4) and satisfying the condition that the segment of any tangent is bisected at the point of tangency can be derived as follows:

1. Let the equation of the conic section be Ax² + By² + 2Gx + 2Fy + C = 0.
2. Since the segment of any tangent is bisected at the point of tangency, the midpoint of the segment is also on the conic section. Therefore, the equation of the midpoint can be written as (x', y') = (2, 4).
3. Substituting the values of x' and y' into the equation of the conic section, we get A(2)² + B(4)² + 2G(2) + 2F(4) + C = 0.
4. Simplifying the equation, we have 4A + 16B + 4G + 8F + C = 0.
5. Since the equation of an auxiliary circle is of the form x² + y² = r², we can compare the coefficients of the equation of the auxiliary circle with the simplified equation of the conic section. This gives us the equation 4A + 16B + 4G + 8F + C = 0.
6. Therefore, the equation of the auxiliary circle is x² + y² = 4√2.