Question

Prove that the curves xy = 4 and x² + y² = 8 touch each other.

Answer 1

Given circles are xy = 4   .....(i)

And x² + y² = 8   .....(ii)

Differentiating equation (i) w.r.t., x

${\displaystyle x\cdot \frac{\text{dy}}{\text{dx}}+y\cdot 1}$ = 0

⇒ ${\displaystyle \frac{\text{dy}}{\text{dx}}=-\frac{y}{x}}$

⇒ m₁ = ${\displaystyle -\frac{y}{x}}$  .....(iii)

Where, m₁ is the slope of the tangent to the curve.

Differentiating equation (ii) w.r.t. x

${\displaystyle 2x+2y\cdot \frac{\text{dy}}{\text{dx}}}$ = 0

⇒ ${\displaystyle \frac{\text{dy}}{\text{dx}}=-\frac{x}{y}}$

⇒ m₂ = ${\displaystyle -\frac{x}{y}}$

Where, m₂ is the slope of the tangent to the circle.

To find the point of contact of the two circles

m₁ = m₂

⇒ ${\displaystyle -\frac{y}{x}=-\frac{x}{y}}$

⇒  x² = y²

Putting the value of y² in equation (ii)

x² + x² = 8

⇒ 2x² = 8

⇒ x² = 4

∴ x = ± 2

∵ x2 = y2

⇒ y = ± 2

∴ The point of contact of the two circles are (2, 2) and (– 2, 2).