Let a variable line passing through the centre of the circle x^2+y^2-16 x-4 y=0 , meet the positive co-ordinate axes at the points A and B . Then the minimum value of O A+O B , where O is the origin, is equal to

Circle PYQ — JEE Main 2024 · 31 Jan · Shift 2 · Q48 with Solution

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Correct answer: D

Given circle: $x^2+y^2-16x-4y=0$

Complete the squares: $x^2-16x+y^2-4y=0$ $x^2-16x+64+y^2-4y+4=68$ $(x-8)^2+(y-2)^2=68$

So the centre is $C=(8,2).$

Let the line meet the positive $x$ -axis at $A(a,0)$ and the positive $y$ -axis at $B(0,b)$ , where $a>0,\, b>0$ .

Its intercept form is $\frac{x}{a}+\frac{y}{b}=1.$

Since it passes through the centre $(8,2)$ , $\frac{8}{a}+\frac{2}{b}=1.$

We need to minimize $OA+OB=a+b.$

From $\frac{8}{a}+\frac{2}{b}=1,$ write it as $\frac{8}{a}=1-\frac{2}{b} = \frac{b-2}{b}$ so $a=\frac{8b}{b-2}, \qquad b>2.$

Then $S=a+b=\frac{8b}{b-2}+b.$

Simplify:

=\frac{b^2+6b}{b-2}.$$ Now divide: $$\frac{b^2+6b}{b-2}=b+8+\frac{16}{b-2}.$$ Let $$t=b-2>0.$$ Then $$S=(t+2)+8+\frac{16}{t}=t+10+\frac{16}{t}.$$ --- 4. **Minimize** By AM-GM, $$t+\frac{16}{t} \ge 2\sqrt{16}=8.$$ Hence $$S=t+10+\frac{16}{t} \ge 10+8=18.$$ Equality holds when $$t=4 \Rightarrow b-2=4 \Rightarrow b=6.$$ Then $$a=\frac{8\cdot 6}{6-2}=12.$$ So the minimum value is $$a+b=12+6=18.$$ --- 5. **Check options** The minimum value of $OA+OB$ is $$18.$$ So the correct option is: **D**.