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Statement 1 : The only circle having radius $\sqrt{10}$ and a diameter along line $2 x+y=5 \, is\, x^{2}+y^{2}-6x+2y=0.$
Statement 2 : $2x + y = 5$ is a normal to the circle $x^{2}+y^{2}-6x+2y=0.$
Statement 1 : The only circle having radius $\sqrt{10}$ and a diameter along line $2 x+y=5 \, is\, x^{2}+y^{2}-6x+2y=0.$
Statement 2 : $2x + y = 5$ is a normal to the circle $x^{2}+y^{2}-6x+2y=0.$
Updated On: Sep 24, 2024
- Statement 1 is false; Statement 2 is true
- Statement 1 is true; Statement 2 is true, Statement 2 is a correct explanation for Statement 1
- Statement 1 is true; Statement 2 is false
- Statement 1 is true; Statement 2 is true; Statement 2 is not a correct explanation for Statement 1
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The Correct Option is A
Solution and Explanation
Circle: $x^{2}+y^{2}-6x+2y=0$ ...(i)
Line : $2\times + 3 - 1 = 5$ ...(ii)
Centre=(3,-1)
Now, $2 \times 3 -1 = 5$, hence centre lies on the given line. Therefore line passes through the centre. The given line is normal to the circle.
Thus statement-2 is true, hut statement-1 is not true as there are infinite circle according to the given conditions.
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