JEE Main 2013 (Online) 25th AprilQuestion 1
Statement 1 : The only circle having radius $\sqrt{10}$ and a diameter along line $2x+y=5$ is $x^{2}+y^{2}-6x+2y=0$.
Statement 2 : $2x+y=5$ is a normal to the circle $x^{2}+y^{2}-6x+2y=0$.
Circle
Solution
Concept. A line is a normal to a circle $\iff$ it passes through the centre. Formulas used. (1) Centre of $x^2+y^2+2gx+2fy+c=0$ is $(-g,-f)$.Step 1 - Statement 2. Centre $(3,-1)$: $2(3)+(-1)=5$. The line passes through the centre, so it is a normal. True. Step 2 - Statement 1. A circle of radius $\sqrt{10}$ with a diameter along $2x+y=5$ only needs its centre on that line. $(3,-1)$ works, but so does $(0,5)$: $(x)^2+(y-5)^2=10$. So the circle is not unique. False. Answer: A