Question

The centre of a circle passing through the point (0, 1) and touching the curve y = x² at (2, 4), is-

• A. $\left( \frac{- 16}{5},\frac{27}{10} \right)$
• B. $\left( \frac{- 16}{7},\frac{5}{10} \right)$
• C. $\left( \frac{- 16}{5},\frac{53}{10} \right)$
• D. None

Answer 1

Answer: (C)

$\left( \frac{- 16}{5},\frac{53}{10} \right)$

Answer 2

Equation of the tangent to the curve y = x² at (2, 4) is

$\frac{y + 4}{2}$= 2x i.e. 4x – y – 4 = 0

Equation of any circle touching the curve y = x² at (2, 4) can be written as

(x – 2)² + (y – 4)² + l(4x – y – 4) = 0

Since the circle passes through (0, 1), therefore

2² + 3² + l (–1 – 4) = 0

Gives l = $\frac{13}{5}$

Hence, equation the circle is

x² + y² + (4l – 4)x – (l + 8)y + 20 – 4l = 0

whose centre C ŗ$\left( 2 - 2\lambda,\frac{\lambda}{2} + 4 \right)$

ŗ $\left( 2 - \frac { 26 } { 5 } , \frac { 13 } { 10 } + 4 \right)$ ŗ$\left( \frac{- 16}{5},\frac{53}{10} \right)$.