JEE PYQ: Circle Question 2
Question 2 - 2021 (16 Mar Shift 2)
Let the lengths of intercepts on $x$-axis and $y$-axis made by the circle $x^2 + y^2 + ax + 2ay + c = 0$, $(a < 0)$ be $2\sqrt{2}$ and $2\sqrt{5}$, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line $x + 2y = 0$, is equal to:
(1) $\sqrt{11}$
(2) $\sqrt{7}$
(3) $\sqrt{6}$
(4) $\sqrt{10}$
Show Answer
Answer: (3)
Solution
Center $(-a/2, -a)$, $r^2 = a^2/4 + a^2 - c$. $x$-intercept $= 2\sqrt{a^2/4 - c} = 2\sqrt{2}$ gives $a^2/4 - c = 2$. $y$-intercept $= 2\sqrt{a^2 - c} = 2\sqrt{5}$ gives $a^2 - c = 5$. So $a = -2$ (since $a < 0$), $c = -1$. Circle: $(x-1)^2 + (y-2)^2 = 6$. Tangent perpendicular to $x+2y=0$ has slope 2. Distance from origin to tangent $= \frac{|2(1) - 2 \pm \sqrt{30}|}{\sqrt{5}} = \sqrt{6}$.
BETA
