Question

Point on the line y = 2x + 11 that is nearest to the circle

16x² + 16y² + 32x - 8y − 50 = 0, is

• A. $\left( - \frac { 7 } { 2 } , 4 \right)$
• B. $\left( - \frac { 9 } { 2 } , 2 \right)$
• C. (-3, 5)
• D. ) $\left( \frac { 1 } { 2 } , 12 \right)$

Answer 1

Answer: (B)

$\left( - \frac { 9 } { 2 } , 2 \right)$

Answer 2

Clearly 'A' is the required point. Equation of line ABB₁ is

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

Solving it with y = 2x + 11 we get A ≡ $\left( \frac { 9 } { 2 } , 2 \right)$