Question

The point on the straight line \(y = 2x + 11\) which is nearest to the circle \(16(x^2 + y^2) + 32x - 8y - 50 = 0\), is

• A. \(\left(\frac{9}{2},2\right)\)
• B. \(\left(\frac{9}{2},-2\right)\)
• C. \(\left(-\frac{9}{2},2\right)\)
• D. \(\left(-\frac{9}{2},-2\right)\)

Hint:

Nearest point from a line to a circle is always the foot of perpendicular from the center to the line.

Answer 1

The Correct Option is C

Concept: Nearest point on a line to a circle lies along the perpendicular from the center of the circle.

Step 1:Convert circle to standard form
\[ 16(x^2 + y^2) + 32x - 8y - 50 = 0 \] Divide by 16: \[ x^2 + y^2 + 2x - \frac{1}{2}y - \frac{25}{8} = 0 \] \[ (x+1)^2 + \left(y-\frac{1}{4}\right)^2 = \text{constant} \] \[ \Rightarrow \text{Center } C = (-1,\, \tfrac{1}{4}) \]

Step 2:Line equation \[ y = 2x + 11 \Rightarrow 2x - y + 11 = 0 \]

Step 3:Foot of perpendicular from center \[ x' = x_1 - \frac{a(ax_1 + by_1 + c)}{a^2+b^2}, \quad y' = y_1 - \frac{b(ax_1 + by_1 + c)}{a^2+b^2} \] Here \(a=2, b=-1, c=11\), \((x_1,y_1)=(-1,\tfrac{1}{4})\) \[ ax_1 + by_1 + c = -2 - \frac{1}{4} + 11 = \frac{35}{4} \] \[ a^2 + b^2 = 5 \] \[ x' = -1 - \frac{2 \cdot \frac{35}{4}}{5} = -1 - \frac{70}{20} = -\frac{9}{2} \] \[ y' = \frac{1}{4} - \frac{-1 \cdot \frac{35}{4}}{5} = \frac{1}{4} + \frac{35}{20} = 2 \] Conclusion \[ {\left(-\frac{9}{2},\,2\right)} \]