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. $(\frac{9}{2},2)$
• B. $(-\frac{9}{2},2)$
• C. $(\frac{9}{2},-2)$
• D. none of these

Answer 1

The correct option is B $(-\frac{9}{2},2)$

The given straight line doesn't cut the circle
The centre of the circle is $(-1,\frac{1}{4})$
Shortest distance is the distance between parallel lines $y=2x+11$and $y=2x+h$ (tangent to the circle)
Let (k,2k+11) be the general point on the line.
As the tangent and normal are perpendicular $\left(\frac{2k+11-\frac{1}{4}}{k+1}\right)\left(2\right)=-1$

$\Rightarrow k=\frac{-9}{2}$
Hence the required point becomes $(-\frac{9}{2},2)$