Question

Prove that the curves y² = 4x and x² + y² – 6x + 1 = 0 touch each other at the point (1, 2)

Answer 1

Given that the equation of the two curves are y² = 4x  .....(i)

And x² + y² – 6x + 1 = 0   .....(ii)

Differentiating (i) w.r.t. x, we get `2y  "dy"/"dx"` = 4

⇒ `"dy"/"dx" = 2/y`

Slope of the tangent at (1, 2)

m₁ = `2/2` = 1

Differentiating (ii) w.r.t. x

⇒ `2x + 2y * "dy"/"dx" - 6` = 0

⇒ `2y * "dy"/"dx"` = 6 – 2x

⇒ `"dy"/"dx" = (6 - 2x)/(2y)`

∴ Slope of the tangent at the same point (1, 2)

⇒ m₂ = `(6 - 2 xx 1)/(2 xx 2)`

= `4/4`

= 1

We see that m₁ = m₂ = 1 at the point (1, 2).

Hence, the given circles touch each other at the same point (1, 2).