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

Let $P$ be the point on the line

$2x-y+11=0$ ...(1)

which is nearest to the circle $x^2+y^2+2x-\frac{1}{2}y-\frac{25}{8}=0$ ...(2)

with centre $C(-1,\frac{1}{4}).$

Then, $CP$ is $\perp$ to the line (1) and $CP>$ radius.

[Note that if $CP\le r,$ the line intersects or touches the circle and then the point of intersection or point of contact are required points]

Here, $CP=\frac{|-2-\frac{1}{4}+11|}{\sqrt{5}}=\frac{35}{4\sqrt{5}}>\frac{\sqrt{67}}{4}$ (radius).

Now, equation of $CP[\perp$ to line (1)$]$ is

$x+2y=\lambda ,$ where $1+\frac{1}{2}=\lambda$ or $\lambda =-\frac{1}{2}.$

$\therefore$ Equation of $CP$ is $2x+4y+1=0$ ...(3)

Solving (1) and (3), we get $y=2,x=-\frac{9}{2}.$

Hence, the required point is $(\frac{-9}{2},2).$