Prove that the pair of straight lines joining the origin to the points of inersection of the circles x² + y² = a² and x² + y² + 2(gx + fy) = 0 is a² (x² + y²) − 4(gx + fy)² = 0. -
Prove that the pair of straight lines joining the origin to the points of inersection of the circles x² + y² = a² and x² + y² + 2(gx + fy) = 0 is a² (x² + y²) − 4(gx + fy)² = 0. -
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Prove that the pair of straight lines joining the origin to the points of
inersection of the circles x² + y² = a² and x² + y² + 2(gx + fy) = 0 is
a² (x² + y²) - 4(gx+fy)² = 0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F80c1fede-b0f4-4a7c-b099-9c7972b6dfa9%2Fc9838bf9-b2b2-4c73-909e-d1d6b8aa66ac%2Faw02hd_processed.png&w=3840&q=75)
Transcribed Image Text:Prove that the pair of straight lines joining the origin to the points of
inersection of the circles x² + y² = a² and x² + y² + 2(gx + fy) = 0 is
a² (x² + y²) - 4(gx+fy)² = 0.
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