At what point on the circle x2 + y2 – 2x – 4y + 1 = 0, the tangent

1.	At what point on the circle x2 + y2 – 2x – 4y + 1 = 0, the tangent is parallel to x – axis.
Answer» Given: The curve is x² + y² – 2x – 4y + 1 = 0 Differentiating the above w.r.t x ⇒ x² + y² – 2x – 4y + 1 = 0 ⇒ 2x^{2 – 1} + 2y^{2 – 1} x \(\frac{dy}{dx}\)– 2 – 4 x \(\frac{dy}{dx}\)+ 0 = 0 ⇒ 2x + 2y \(\frac{dy}{dx}\)– 2 – 4\(\frac{dy}{dx}\)= 0 ⇒ \(\frac{dy}{dx}\) (2y – 4) = – 2x + 2 ⇒ \(\frac{dy}{dx}\) \(= \frac{-2(x-1)}{2(y-2)}\) ⇒ \(\frac{dy}{dx}\) \(= \frac{-(x-1)}{(y-2)}\)......(1) \(\therefore\) \(\frac{dy}{dx}\) = The Slope of the tangent = tanθ Since, the tangent is parallel to x – axis i.e, ⇒ \(\frac{dy}{dx}\) = tan(0) = 0 ...(2) \(\therefore\) tan(0) = 0 From (1) & (2),we get, ⇒\(=\frac{-(x-1)}{(y-2)}=0\) ⇒ – (x – 1) = 0 ⇒ x = 1 Substituting x = 1 in x² + y² – 2x – 4y + 1 = 0,we get, ⇒ 1² + y² – 2 x1 – 4y + 1 = 0 ⇒ 1 – y² – 2 – 4y + 1 = 0 ⇒ y² – 4y = 0 ⇒ y(y – 4) = 0 ⇒ y = 0 & y = 4 Thus, the required point is (1,0) & (1,4)

At what point on the circle x2 + y2 – 2x – 4y + 1 = 0, the tangent is parallel to x – axis.

Answer»

Given:

The curve is x² + y² – 2x – 4y + 1 = 0

Differentiating the above w.r.t x

⇒ x² + y² – 2x – 4y + 1 = 0

⇒ 2x^{2 – 1} + 2y^{2 – 1} x \(\frac{dy}{dx}\)– 2 – 4 x \(\frac{dy}{dx}\)+ 0 = 0

⇒ 2x + 2y \(\frac{dy}{dx}\)– 2 – 4\(\frac{dy}{dx}\)= 0

⇒ \(\frac{dy}{dx}\) (2y – 4) = – 2x + 2

⇒ \(\frac{dy}{dx}\) \(= \frac{-2(x-1)}{2(y-2)}\)

⇒ \(\frac{dy}{dx}\) \(= \frac{-(x-1)}{(y-2)}\)......(1)

\(\therefore\) \(\frac{dy}{dx}\) = The Slope of the tangent = tanθ

Since, the tangent is parallel to x – axis

i.e,

⇒ \(\frac{dy}{dx}\) = tan(0) = 0 ...(2)

\(\therefore\) tan(0) = 0

From (1) & (2),we get,

⇒\(=\frac{-(x-1)}{(y-2)}=0\)

⇒ – (x – 1) = 0

⇒ x = 1

Substituting x = 1 in x² + y² – 2x – 4y + 1 = 0,we get,

⇒ 1² + y² – 2 x1 – 4y + 1 = 0

⇒ 1 – y² – 2 – 4y + 1 = 0

⇒ y² – 4y = 0

⇒ y(y – 4) = 0

⇒ y = 0 & y = 4

Thus, the required point is (1,0) & (1,4)