Question

Equation of circle of minimum radius which touches both the parabolas y = x² + 2x + 4 & x = y² + 2y + 4 is –

• A. 4x² + 4y² – 11x – 11y – 13 = 0
• B. 2x² + 2y² – 11x – 11y – 13 = 0
• C. 3x² + 3y² – 11x – 11y – 13 = 0
• D. x² + y² – 11x – 11y – 13 = 0

Answer 1

Answer: (A)

4x² + 4y² – 11x – 11y – 13 = 0

Answer 2

y = x² + 2x + 4

$\frac{dy}{dx}$ = 2x + 2 = 1 \ x = – $\frac{1}{2}$

y = $\frac{1}{4}$ – 1 + 4 = $\frac{13}{4}$

\ point on y = x² + 2x + 4 is $\left( - \frac{1}{2},\frac{13}{4} \right)$

corresponding point on x = y² + 2y + 4 is $\left( \frac{13}{4},\frac{- 1}{2} \right)$

\ equation of circle is

$\left( x + \frac{1}{2} \right)$ $\left( x - \frac{13}{4} \right)$+ $\left( y - \frac { 13 } { 4 } \right)$ $\left( y + \frac{1}{2} \right)$ = 0

i.e. x² + y² – $\frac{11}{4}$x – $\frac{11}{4}$y – $\frac{13}{4}$ = 0

i.e. 4x² + 4y² – 11x – 11y – 13 = 0