ax+by=cx2+y2=1Prove that bx-ay= +- Root Over(a2 + b2 + c2)

1.	ax+by=cx2+y2=1Prove that bx-ay= +- Root Over(a2 + b2 + c2)
Answer» Given that ax + by = c ... (1) x² + y² = 1 ... (2) Squaring (1) on both sides, we get ⇒ a²x² + b²y² + 2axby = c² ⇒ a²(1 - y²) + b²(1 - x²) + 2axby = c² (∵ x² + y² = 1) ⇒ a² - a²y² + b² - b²x² + 2axby = c² ⇒ a² + b² - c² = a²y² + b²x² - 2axby ⇒ a² + b² - c² = (bx - ay)² ∴ bx - ay = \( { \pm \sqrt{a^2 + b^2 - c^2} }\)

Answer»

Given that

ax + by = c ... (1)

x² + y² = 1 ... (2)

Squaring (1) on both sides, we get

⇒ a²x² + b²y² + 2axby = c²

⇒ a²(1 - y²) + b²(1 - x²) + 2axby = c² (∵ x² + y² = 1)

⇒ a² - a²y² + b² - b²x² + 2axby = c²

⇒ a² + b² - c² = a²y² + b²x² - 2axby

⇒ a² + b² - c² = (bx - ay)²

∴ bx - ay = \( { \pm \sqrt{a^2 + b^2 - c^2} }\)

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ax+by=cx2+y2=1Prove that bx-ay= +- Root Over(a2 + b2 + c2)

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